Answer:
Explanation:
C is center of mass of the bar ie middle point. Spring is attached with one end ie A of the bar . When this end is displaced by distance x ( small) a restoring force kx is produced which creates a torque
Torque created = kx . l /2 , which creates angular acceleration α
moment of inertia I = ml²/12
torque = I. x α
kx . L /2 = I. x α
α = L kx / 2I
a = linear acceleration
= α x L/2
= L² kx / 4I
= L²x 12 kx / 4 mL²
a = 3kx/m
This shows that motion is SHM as acceleration is proportional to displacement x .
angular frequency ω² = 3k / m
frequency f = 1/2π √ 3k/m
= (1/6.28) x √ 3x 500/8
= 2.18 Hz
Final answer:
The frequency of small oscillations is approximately 3.98 Hz, and the smallest value of the spring constant for which oscillations will occur is 0 N/m.
Explanation:
(a) To determine the frequency of small oscillations, we can use the formula: f = (1/2π) × (√(k/m)), where f is the frequency, k is the spring constant, and m is the mass of the bar. Plugging in the values, we get:
f = (1/2π) × (√(500/8)) ≈ 3.98 Hz
(b) To find the smallest value of the spring constant k for which oscillations will occur, we need to consider critical damping. Critical damping occurs when the damping force is equal to the force exerted by the spring. The formula for critical damping is c = 2√(mk), where c is the damping coefficient, m is the mass, and k is the spring constant. Since we know that the damping coefficient is 0, we can solve for k to get:
k = (c^2)/(4m). Substituting in the values, we get:
k = (0^2)/(4 x 8) = 0 N/m
Assume that the home construction industry is perfectly competitive and in long-run competitive equilibrium. It follows that: A. marginal cost exceeds long-run average total cost.B. marginal cost equals long-run average total cost.C. there will be incentive for new firms to enter the industry.D. firms in the industry enjoy economic profits.
Answer:
B. Marginal cost equals long-run average total cost.
Explanation:
The zero profit condition implies that entry continues until all firms are producing at minimum long run average total cost. Since the marginal cost curve cuts the long run average total cost curve at its minimum point, marginal cost and long run average total cost must be equal in long run equilibrium.
A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.90×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.
Answer:
a. Δx = 2.59 cm
Explanation:
mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m
Using momentum conserved
mb * (0) + mp * vp = ( mb + mp ) * vf
vf = ( mp / mp + mb) * vp
¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²
Solve to Δx '
Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )
Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97 m /s) ² / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]
Δx = 0.02599 m ⇒ 2.59 cm
When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is 278 nm. What is the vacuum wavelength of the light in nm?
Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.
From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as
[tex]2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}[/tex]
We are given the second smallest nonzero thickness at which destructive interference occurs.
This corresponds to, m = 2, therefore
[tex]2t = 2\lambda_{film}[/tex]
[tex]t = \lambda_{film}[/tex]
The index of refraction of soap is given, then
[tex]\lambda_{film} = \frac{\lambda_{vacuum}}{n}[/tex]
Combining the results of all steps we get
[tex]t = \frac{\lambda_{vacuum}}{n}[/tex]
Rearranging, we find
[tex]\lambda_{vacuum} = tn[/tex]
[tex]\lambda_{vacuum} = (278)(1.33)[/tex]
[tex]\lambda_{vacuum} = 369.74nm[/tex]
According to the videos seen in Modules 8, please write a short paragraph answering the following questions: a. What are some examples of on-the-spot feedback? b. What are some examples of delayed feedback?
Answer:
.
Explanation:
How much energy will an electron gain if it moves through a potential difference of 1.0 V?
a. 1.0 eV
b. 0.5 eV
c. 2.0 eV
d. 1.6 x 10-19 eV
Answer:
a. 1.0 eV
Explanation:
Given that
Voltage difference ,ΔV = 1 V
From work power energy
Work =Change in the kinetic energy
We know that work on the charge W= q ΔV
For electron ,e= 1.6 x 10⁻¹⁹ C
q=e= 1 x 1.6 x 10⁻¹⁹ C
Change in the kinetic energy
ΔKE= q ΔV
Now by putting the values
ΔKE= 1 x 1.6 x 10⁻¹⁹ x 1 C.V
We can also say that
ΔKE= 1 e.V
Therefore the answer will be a.
a). 1.0 eV
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. Suppose the skater has a mass of 64 kg . One eighth of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. The rest of her mass is approximately in the form of a 20-cm-radius cylinder.A. Estimate the skater's moment of inertia to two significant figures.B. If she were to hold her arms outward, rather than at her sides, would her moment of inertia increase, decrease, or remain unchanged?
Answer:
Explanation:
The body can be split into two parts
1 ) cylinder part of mass 7/8 x 64 = 56 kg of radius 20 cm
2 ) In the form of rod attached with cylinder having length 40 cm and mass
of 1/8 x 64 = 8kg
moment of inertia of cylinder
= 1/2 mr²
= .5 x 56 x ( 20 x 10⁻²)²
= 1.12 kg m²
moment of inertia of rods
= 1/3 ml²
= 1/3 x 8 x ( 40 x 10⁻²)²
= .4267 kg m²
Total MI = 1.5467 Kg m²
A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?
Answer:
22.1 m
Explanation:
[tex]v_{o}[/tex] = initial speed of ball = 14.3 m/s
[tex]\theta[/tex] = Angle of launch = 27°
Consider the motion of the ball along the vertical direction.
[tex]v_{oy}[/tex] = initial speed of ball = [tex]v_{o} Sin\theta = 14.3 Sin27 = 6.5 ms^{-1}[/tex]
[tex]a_{y}[/tex] = acceleration due to gravity = - 9.8 ms⁻²
[tex]t[/tex] = time of travel
[tex]y[/tex] = vertical displacement = - 3.50 m
Using the kinematics equation that suits the above list of data, we have
[tex]y = v_{oy} t + (0.5) a_{y} t^{2} \\- 3.50 = (6.5) t + (0.5) (- 9.8) t^{2}\\- 3.50 = (6.5) t - 4.9 t^{2}\\t = 1.74 s[/tex]
Consider the motion of the ball along the horizontal direction.
[tex]v_{ox}[/tex] = initial speed of ball = [tex]v_{o} Cos\theta = 14.3 Cos27 = 12.7 ms^{-1}[/tex]
[tex]X[/tex] = Horizontal distance traveled
[tex]t[/tex] = time taken = 1.74 s
Since there is no acceleration along the horizontal direction, we have
[tex]X = v_{ox} t\\X = (12.7)(1.74)\\X = 22.1 m[/tex]
Use Hooke's Law for this (F = - k s ): Where F is the spring's restoring force; k is the spring constant; and s is the stretch. The negative sign means the spring's restoring force is opposite the stretch direction. You have a plot from weight [N] versus stretch [m]. The data forms a linear trend y = 3.662 * x + 1.67. How much will the spring stretch if 51.7 grams is hung on the spring? Answer in centimeters with three significant figures or N/A if not enough information is given to answer. When you calculate your ansswer, don't use the negative sign in the Hooke's Law formula. Just know that the negative sign simply denotes the force direction is opposite the stretch (or compression).
The spring will stretch by 13.79 cm when 51.7 grams is hung on it.
Explanation:To find the stretch of the spring when a weight of 51.7 grams is hung on it, we will use Hooke's Law. First, we need to convert the weight to Newtons. Since 1 g is equal to 0.0098 N, the weight in Newtons is 51.7 grams * 0.0098 N/g = 0.50546 N. Now we can rearrange Hooke's Law equation, F = -k * s, to solve for s, the stretch of the spring. Plugging in the values, we get 0.50546 N = -k * s. Rearranging further, we have s = 0.50546 N / -3.662 = -0.1379 m. Since the question asks for the answer in centimeters, we can convert -0.1379 m to centimeters by multiplying by 100. Therefore, the spring will stretch by 13.79 cm when 51.7 grams is hung on it.
A 1.53-kg piece of iron is hung by a vertical ideal spring. When perturbed slightly, the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. If we choose the total potential energy (elastic and gravitational) to be zero at the equilibrium position of the hanging iron, what is the total mechanical energy of the system?
A. 0.844 JB. 0.646 JC. 0.000 JD. 0.955 JE. 0.633 J
To solve this problem it is necessary to apply the concepts related to the elastic potential energy from the simple harmonic movement.
Said mechanical energy can be expressed as
[tex]E = \frac{1}{2} kA^2[/tex]
Where,
k = Spring Constant
A = Cross-sectional Area
From the angular movement we can relate the angular velocity as a function of the spring constant and the mass in order to find this variable:
[tex]\omega^2 = \frac{k}{m}[/tex]
[tex]k = m\omega^2\rightarrow \omega = 2\pi f [/tex] for f equal to the frequency.
[tex]k = 1.53(2\pi 1.95)^2[/tex]
[tex]k = 229.44[/tex]
Finally the energy released would be
[tex]E = \frac{1}{2} (229.44)(0.075)^2[/tex]
[tex]E = 0.6453J \approx 0.646J[/tex]
Therefore the correct answer is B.
The correct answer is option A. The total mechanical energy of the system is [tex]\( 0.844 \, \text{J} \)[/tex]
To find the total mechanical energy of the system, we need to calculate both the elastic potential energy (due to the spring) and the gravitational potential energy (due to the height above the equilibrium position) at the maximum amplitude point.
First, let's calculate the spring constant [tex]\( k \)[/tex] using the formula for the frequency[tex]\( k \)[/tex] of a mass-spring system:
[tex]\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \][/tex]
where \( m \) is the mass of the iron piece. Solving for [tex]\( k \)[/tex], we get:
[tex]\[ k = (2\pi f)^2 m \][/tex]
[tex]\[ k = (2\pi \times 1.95 \, \text{Hz})^2 \times 1.53 \, \text{kg} \][/tex]
[tex]\[ k \approx 244.73 \, \text{N/m} \][/tex]
Next, we calculate the elastic potential energy [tex]\( U_{\text{elastic}} \)[/tex] at the maximum amplitude [tex]\( A \)[/tex]:
[tex]\[ U_{\text{elastic}} = \frac{1}{2} k A^2 \][/tex]
[tex]\[ U_{\text{elastic}} = \frac{1}{2} \times 244.73 \[/tex], [tex]\text{N/m} \times (0.0750 \[/tex], [tex]\text{m})^2 \] \[ U_{\text{elastic}} \approx 0.683 \[/tex], [tex]\text{J} \][/tex]
Now, let's calculate the gravitational potential energy [tex]\( U_{\text{gravitational}} \)[/tex]at the maximum amplitude:
[tex]\[ U_{\text{gravitational}} = m g h \][/tex]
where [tex]\( g \)[/tex]is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \[/tex])) and [tex]\( h \)[/tex] is the height at maximum amplitude. Since the amplitude is given in centimeters, we convert it to meters:
[tex]\[ h = 7.50 \, \text{cm} = 0.0750 \, \text{m} \][/tex]
[tex]\[ U_{\text{gravitational}} = 1.53 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.0750 \, \text{m} \][/tex]
[tex]\[ U_{\text{gravitational}} \approx 1.130 \, \text{J} \][/tex]
The total potential energy [tex]\( U_{\text{total}} \)[/tex] at the maximum amplitude is the sum of the elastic and gravitational potential energies:
[tex]\[ U_{\text{total}} = U_{\text{elastic}} + U_{\text{gravitational}} \] \[ U_{\text{total}} \approx 0.683 \, \text{J} + 1.130 \, \text{J} \] \[ U_{\text{total}} \approx 1.813 \, \text{J} \][/tex]
However, we must consider that the total potential energy is zero at the equilibrium position. Therefore, the total mechanical energy [tex]\( E \)[/tex] of the system is equal to the total potential energy at the maximum amplitude:
[tex]\[ E = U_{\text{total}} \] \[ E \approx 1.813 \, \text{J} \][/tex]
Since the potential energy at the equilibrium position is zero, the total mechanical energy is simply the sum of the elastic and gravitational potential energies at the maximum amplitude, which is approximately [tex]\( 1.813 \, \text{J} \)[/tex]. However, the options given are in the range of [tex]\( 0.633 \, \text{J} \) to \( 0.955 \, \text{J} \)[/tex], which suggests that there might be a mistake in the calculation or interpretation of the problem.
Upon re-evaluating the calculation, it seems that the gravitational potential energy was incorrectly calculated. The correct calculation should consider that the gravitational potential energy is zero at the equilibrium position, not at the maximum amplitude. Therefore, we should only consider the change in gravitational potential energy from the equilibrium to the maximum amplitude, which is half the amplitude squared divided by the spring constant, as the mass will oscillate symmetrically around the equilibrium position.
The correct gravitational potential energy [tex]\( U_{\text{gravitational}} \)[/tex] at the maximum amplitude is:
[tex]\[ U_{\text{gravitational}} = \frac{1}{2} k \left(\frac{h}{2}\right)^2 \][/tex]
[tex]\[ U_{\text{gravitational}} = \frac{1}{2} \times 244.73 \, \text{N/m} \times \left(\frac{0.0750 \, \text{m}}{2}\right)^2 \][/tex]
[tex]\[ U_{\text{gravitational}} \approx 0.160 \, \text{J} \][/tex]
Now, the total mechanical energy [tex]\( E \)[/tex] of the system is:
[tex]\[ E = U_{\text{elastic}} + U_{\text{gravitational}} \] \[ E \approx 0.683 \, \text{J} + 0.160 \, \text{J} \] \[ E \approx 0.844 \, \text{J} \][/tex]
A 2-kg object is initially at the bottom of a long 50° inclined plane, and is beginning to slide up this inclined plane. The initial velocity of the object os 3.0 m/s. The coefficient of kinetic friction between the object and the surface of the inclined plane is µk = 0.3, while the coefficient of static friction is µs = 0.4. Will the object ever slide back down to the initial position? If yes, what will be its speed once it slides back down to the bottom? If no, how high up the incline will it manage to climb?
Answer:
Explanation:
Given
mass of object [tex]m=2 kg[/tex]
inclination [tex]\theta =50^{\circ}[/tex]
[tex]\mu _k=0.3[/tex]
[tex]\mu _s=0.4[/tex]
initial velocity [tex]u=3 m/s[/tex]
acceleration of block during upward motion
[tex]a=g\sin \theta -\mu _kg\cos \theta [/tex]
[tex]a=g(\sin 50-0.3\cos 50)[/tex]
[tex]a=5.617 m/s^2[/tex]
using relation
[tex]v^2-u^2=2a\cdot s[/tex]
where [tex]s=distance\ moved [/tex]
[tex]v=final\ velocity[/tex]
v=0 because block stopped after moving distance s
[tex]0-(3)^2=2\cdot (-5.617)\cdot s[/tex]
[tex]s=\frac{4.5}{5.617}[/tex]
[tex]s=0.801[/tex]
If block stopped after s m then force acting on block is
[tex]F=mg\sin \theta =[/tex]friction force [tex]f_r=\mu mg\cos \theta [/tex]
[tex]F>f_r[/tex] therefore block will slide back down to the bottom
Which pair of terms correctly fills the blanks in the following sentence? ________ is a property of an object while ________ is a property of a material.
A) resistivity, conductivityB) current, resistanceC) resistance, currentD) resistance, resistivity
Answer:
D) resistance, resistivity
Explanation:
Resistance is a physical quantity that indicates the opposition of an object to conduct electricity, this quantity depends on different factors such as temperature, material, object length, among other things. The resistance of two objects of the same material may be different, because it depends on the specifications of the object.
On the other hand, resistivity is a more general quantity, since it is assigned to materials and depends only on the nature of the material and its temperature.
So the resistivity is related to the meterial rather than the object.
The answer is: Resistance is a property of an object while resistivity is a property of a material.
A gas is compressed at a constant pressure of 0.643 atm from 5.46 L to 2.01 L. In the process, 564 J of energy leaves the gas by heat. What is the work done on the gas? Remember: 1 atm = 1.013 × 105 Pa. Answer in units of J. 008 (part 2 of 2) 10.0 points What is the change in its internal energy? Answer in units of J.
Answer:
Part 1)
- 224.6 J
Part 2)
- 339.4 J
Explanation:
[tex]P[/tex] = Constant pressure acting on gas = 0.643 atm = (0.643) (1.013 x 10⁵) Pa = 0.651 x 10⁵ Pa
[tex]V_{i}[/tex] = initial volume = 5.46 L = 0.00546 m³
[tex]V_{f}[/tex] = final volume = 2.01 L = 0.00201 m³
Work done on the gas is given as
[tex]W = P (V_{f} - V_{i})\\W = (0.651\times10^{5}) ((0.00201) - (0.00546))\\W = - 224.6 J[/tex]
Part 2)
[tex]\Delta U[/tex] = Change in the internal energy
[tex]Q[/tex] = Heat energy escaped = - 564 J
Using First law of thermodynamics
[tex]Q = W + \Delta U\\- 564 = - 224.6 + \Delta U\\ \Delta U = - 339.4 J[/tex]
Air enters a turbine operating at steady state at 8 bar, 1400 K and expands to 0.8 bar. The turbine is well insulated, and kinetic and potential energy effects can be neglected. Assuming ideal gas behavior for the air, what is the maximum theoretical work that could be developed by the turbine in kJ per kg of air flow?
To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables
Mathematically this can be determined as
[tex]\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}[/tex]
Where
[tex]T_1 =[/tex]Temperature at inlet of turbine
[tex]T_2 =[/tex] Temperature at exit of turbine
[tex]P_1 =[/tex] Pressure at exit of turbine
[tex]P_2 =[/tex]Pressure at exit of turbine
The steady flow Energy equation for an open system is given as follows:
[tex]m_i = m_0 = m[/tex]
[tex]m(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W[/tex]
Where,
m = mass
[tex]m_i[/tex] = mass at inlet
[tex]m_0[/tex]= Mass at outlet
[tex]h_i[/tex] = Enthalpy at inlet
[tex]h_0[/tex] = Enthalpy at outlet
W = Work done
Q = Heat transferred
[tex]V_i[/tex] = Velocity at inlet
[tex]V_0[/tex]= Velocity at outlet
[tex]Z_i[/tex]= Height at inlet
[tex]Z_0[/tex]= Height at outlet
For the insulated system with neglecting kinetic and potential energy effects
[tex]h_i = h_0 + W[/tex]
[tex]W = h_i -h_0[/tex]
Using the relation T-P we can find the final temperature:
[tex]\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}[/tex]
[tex]\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}[/tex]
[tex]T_2 = 725.126K[/tex]
From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK
So:
[tex]W = h_i -h_0[/tex]
[tex]W = C_p (T_1-T_2)[/tex]
[tex]W = 1.005(1400-725.126)[/tex]
[tex]W = 678.248kJ/Kg[/tex]
Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg
Final answer:
The maximum theoretical work that could be developed by the turbine, assuming ideal gas behavior, is approximately 0.3195 kJ per kg of air flow.
Explanation:
The maximum theoretical work that could be developed by the turbine can be calculated using the ideal gas law equation:
W = Cv(Tin - Tout)
where W is the work done per unit mass, Cv is the specific heat capacity at constant volume, Tin is the inlet temperature, and Tout is the outlet temperature.
First, we need to find the value of Cv for air. For ideal gases, the specific heat capacity at constant volume can be calculated using the equation:
Cv = (R/M)
where R is the gas constant and M is the molar mass of the gas.
In this case, we will use the molar mass of air, which is approximately 28.97 g/mol. The gas constant R is 8.314 J/(mol·K).
Substituting the values into the equation, we get:
Cv = (8.314 J/(mol·K)) / (28.97 g/mol)
Cv = 0.287 J/(g·K)
Now we can calculate the work:
W = (0.287 J/(g·K))(1400 K - 288 K)
W = 0.287 J/(g·K) x 1112 K
W = 319.544 J/g
Finally, we convert the work from Joules to kilojoules:
W = 319.544 J/g * (1 kJ/1000 J)
W ≈ 0.3195 kJ/g
The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 MPa, determine the required minimum wall thickness to the nearest millimeter using (a) the maximum shear stress theory, and (b) maximum distortion energy theory. Apply a factor of safety of 1.5 against yielding.
Answer:
(a) maximum shear stress t=22.5mm
(b)maximum distortion energy t=19.48mm
Explanation:
Ф₁(sigma)=pd/2t
Ф₁=(5×10⁶×1.5)/2t
Ф₁=3.75×10⁶/t
Ф₂=pd/4t
Ф₂=1.875×10⁶/t
(a) According to maximum shear stress theory
|Ф₁|=ФY/1.5
3.75×10⁶/t=250×10⁶/1.5
t=22.5 mm
(b)According to distortion energy theory
Ф₁²+Ф₂²-(Ф₁Ф₂)=(ФY/1.5)²
(3.75×10⁶/t)²+(1.875×10⁶/t)-{(3.75×10⁶/t)(1.875×10⁶/t)}=(250×10⁶/1.5)²
t=19.48mm
A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for this vehicle to maintain a 70 mi/h speed on a 5% upgrade through an air density of 0.002045 slugs/ft3?
Answer:
i put this in the calculator and my answer is 600. hope this helps
Explanation:
A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2, the minimum tractive effort = 220.3 lb.
To find the minimum tractive effort,
vehicle weight is 2500 lb
drag coefficient is 0.35
frontal area 22 ft^2
vehicle speed is 60 mi/hr = 70 ft/sec
gradient - 5%
air density = 0.002045 slugs/ft^3
What is drag force?The force which exerted on a solid body moving in some relation to a fluid by the fluid's movement is known as a drag force.
Drag force is given as,
Fd = [tex]\frac{1}{2}[/tex]CdρAV²
= [tex]\frac{1}{2}[/tex] × 0.35 × 0.002045× 20× 88²
= 55.4 lb.
Force due to vehicle weight,
Fw = 0.01 ( 1+ υ/147)W
= 0.01 ( 1 + 88/147) 2500
= 39.965 lb.
Force due to gradient
Fg = W × g
= 2500 × 0.05
= 125 lb
Minimum tractive effort
F = Fd + Fw +Fg
= 55.4 + 39.965 + 125
= 220.365 lb
Thus, minimum tractive force can be found as 220.365 lb.
Learn more about drag force,
https://brainly.com/question/12774964
#SPJ5
Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have the frequency 440 Hz and the note E should be at 659 Hz . The tuner can determine this by listening to the beats between the third harmonic of the A and the second harmonic of the E.
A tuner first tunes the A string very precisely by matching it to a 440 Hz tuning fork. She then strikes the A and E strings simultaneously and listens for beats between the harmonics. The beat frequency that indicates that the E string is properly tuned is 2.0 Hz
The tuner starts with the tension in the E string a little low, then tightens it. What is the frequency of the E string when she hears four beats per second?
Answer:
Frequency = 658 Hz
Explanation:
The third harmonics of A is,
[tex]f_{A}=(3)(440Hz)=1320Hz[/tex]
The second harmonics of E is,
[tex]f_{E}=(2)(659Hz)=1318Hz[/tex]
The difference in the two frequencies is,
delta_f = 1320 Hz - 1318 Hz = 2 Hz
The beat frequency between the third harmonic of A and the second harmonic of E is,
delta_f = [tex]3f_{A}-2f_{E}[/tex]
[tex]f_{E}=\frac{3f_{A}-delta_f}{2}[/tex]
We have calculate the frequency of the E string when she hears four beats per second, then
delat_f = 4 Hz
[tex]f_{E}=\frac{3(440Hz)-4Hz}{2}[/tex]
[tex]f_{E}=658Hz[/tex]
Hope this helps!
The distance required for a car to come to a stop will vary depending on how fast the car is moving. Suppose that a certain car traveling down the road at a speed of 10 m / s can come to a complete stop within a distance of 20 m . Assuming the road conditions remain the same, what would be the stopping distance required for the same car if it were moving at speeds of 5 m / s , 20 m / s , or 40 m / s ?
Answer:
5 m
80 m
320 m
Explanation:
[tex]v_{o}[/tex] = Initial speed of the car = 10 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d[/tex] = Stopping distance of the car = 20 m
[tex]a[/tex] = acceleration of the car
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 10^{2} + 2 (20) a\\a = - 2.5 ms^{-2}[/tex]
[tex]v_{o}[/tex] = Initial speed of the car = 5 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d'[/tex] = Stopping distance of the car
[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'\\0^{2} = 5^{2} + 2 (- 2.5) d'\\d' = 5 m[/tex]
[tex]v_{o}[/tex] = Initial speed of the car = 20 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d''[/tex] = Stopping distance of the car
[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d''\\0^{2} = 20^{2} + 2 (- 2.5) d''\\d'' = 80 m[/tex]
[tex]v_{o}[/tex] = Initial speed of the car = 40 ms⁻¹
[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹
[tex]d'''[/tex] = Stopping distance of the car
[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²
On the basis of above data, we can use the kinematics equation
[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'''\\0^{2} = 40^{2} + 2 (- 2.5) d'''\\d''' = 320 m[/tex]
A 2510-kg car is coasting up a hill with a 10.0 degree incline at initial speed 20.0 m/s, when a squirrel suddenly darts out into the street ahead of it, and the driver slams on the brakes, causing the car to skid for 20.0 m before the brake is suddenly released and it continues coasting. The coefficient of kinetic friction between the tires and the pavement is 0.500.
Answer
given,
mass of the car = 2510 Kg
angle of inclination = 10°
initial speed = v₁ = 20 m./s
skid length = 20 m
coefficient of friction = 0.5
Using conservation of energy
[tex]\Delta E = \Delta KE + \Delta U[/tex]
[tex]\Delta E = \dfrac{1}{2}mv^2 + m g h[/tex]
h = d sin θ
[tex]\Delta E = \dfrac{1}{2}mv^2- m g (dsin\theta)[/tex]
[tex]\Delta E = \dfrac{1}{2}\times 2510 \times 20^2- 2510\times 9.8 \times (20 sin10^0)[/tex]
[tex]\Delta E =416572.04\ J[/tex]
now, calculating the magnitude of frictional force is equal to
E = F_f d
[tex]F_f = \dfrac{E}{d}[/tex]
[tex]F_f = \dfrac{416572.04}{20}[/tex]
[tex]F_f = 20828\ N[/tex]
An asteroid is in an elliptical orbit around a distant star. At its closest approach, the asteroid is 0.540 AU from the star and has a speed of 54.0 km/s. When the asteroid is at its farthest distance from the star of 32.0 AU, what is its speed (in km/s)? (1 AU is the average distance from the Earth to the Sun and is equal to 1.496 ✕ 1011 m. You may assume that other planets an
Answer:
0.91125 km/s
Explanation:
[tex]v_1[/tex] = Velocity of planet initially = 54 km/s
[tex]r_1[/tex] = Distance from star = 0.54 AU
[tex]v_2[/tex] = Final velocity of planet
[tex]r_2[/tex] = Final distance from star = 32 AU
As the angular momentum of the system is conserved
[tex]mv_1r_1=mv_2r_2\\\Rightarrow v_1r_1=v_2r_2\\\Rightarrow v_2=\dfrac{v_1r_1}{r_2}\\\Rightarrow v_2=\dfrac{54\times 0.54}{32}\\\Rightarrow v_2=0.91125\ km/s[/tex]
When the exoplanet is at its farthest distance from the star the speed is 0.91125 km/s
Cart 111 of mass mmm is traveling with speed v_0v 0 v, start subscript, 0, end subscript in the + x+xplus, x-direction when it has an elastic collision with cart 222 of mass 3m3m3, m that is at rest. What are the velocities of the carts after the collision?
Answer:
In an elastic collision, the momentum and the kinetic energies are conserved.
Momentum:
[tex]\vec{P_i} = \vec{P_f}\\\vec{P}_1 + \vec{P}_2 = \vec{P}_1' + \vec{P}_2'\\m\vec{v_0} + 0 = m\vec{v_1}' + 3m\vec{v_2}}' \\v_0 = v_1 + 3v_2[/tex]
Kinetic energy:
[tex]K_i = K_f\\K_1 + K_2 = K_1' + K_2'\\\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}m{v_1'}^2 + \frac{1}{2}3m{v_2'}^2\\v_0^2 = {v_1'}^2 + 3{v_2'}^2[/tex]
We have two equations and two unknowns:
[tex]v_0 = v_1' + 3v_2'\\v_0^2 = {v_1'}^2 + 3{v_2'}^2\\\\3v_2' = v_0 - v_1'\\3{v_2'}^2 = {v_0}^2 - {v_1'}^2\\\\3{v_2'}^2 = (v_0 - v_1')(v_0 + v_1') = 3{v_2}'(v_1' + v_0)\\\\v_2' = v_1' + v_0\\3v_2' = v_0 - v_1'\\\\4v_2' = 2v_0\\\\v_2' = v_0/2\\v_1' = -v_0/2[/tex]
Explanation:
The first cart hits the second cart at rest and turns back with half its speed.
The second cart starts moving to the right with half the initial speed of the first cart.
Answer:
v1 = -v0/3 ,v2 = 2v0/3
Explanation:
Two long wires hang vertically. Wire 1 carries an upward current of 1.80 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.00 A. A third wire, wire 3, is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force.
(a) Is this situation possible?
Is it possible in more than one way?
(b) Describe the position of wire 3.
(c) Describe the magnitude and direction of the current in wire 3.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
(a) Yes, this situation is possible. No, It is not possible in more than one way.
(b) The position of the wire is at a distance of 16.3 cm, in the direction left of the wire
(c) The magnitude of the current in wire 3 (I3) is 3.267 A and in a direction pointing in the downward direction.
(a)
The diagrammatic illustration of the information provided in the question can be seen in the image attached below.
From the image: suppose the forces on the wire as a result of the two other wires are equal and opposite, then the third wire will experience no net force. Hence, we can say Yes, the situation is possible.
Given that:
I₁ = 1.80 AI₂ = 4.00 AIt will not be possible to have more than one way, this is because the force acting on the other two wires will have opposite directions but they would not have the same equal magnitude.
The force per unit length existing in-between two current-carrying parallel wires can be determined by using the formula:
[tex]\mathbf{\dfrac{F}{l} = \dfrac{\mu_o I_1I_2}{2 \pi r} }[/tex]
where:
[tex]\mathbf{I_1,I_2 }[/tex] = current passing via conductorsr = distance in-between the parallel conductorsSInce F1 = F2, Then:
[tex]\mathbf{\dfrac{\mu_o\times I_1\times I_3\times l}{2 \pi r}= \dfrac{\mu_o\times I_2\times I_3\times l}{2 \pi (r+0.2)} }[/tex]
[tex]\mathbf{\dfrac{I_1}{r}= \dfrac{ I_2}{ (r+0.2)} }[/tex]
[tex]\mathbf{\dfrac{1.80}{r_1}= \dfrac{ 4.00}{ (r_1+0.2)} }[/tex]
1.80(r₁ + 0.2) = 4.00r₁
1.80r₁ +0.36 = 4.00 r₁
0.36 = 2.2r₁
r₁ = 0.36/2.2
r₁ = 16.3 cm in the direction left of the wire.
Again, we are to determine the magnitude of the current and its direction in wire 3 (I₃).
From the image, the forces of F21 = F23
∴
[tex]\mathbf{\dfrac{\mu_o \times I_1 \times I_2 \times l }{2 \pi r} = \dfrac{\mu_o \times I_2 \times I_3 \times l }{2 \pi r}}[/tex]
[tex]\mathbf{\dfrac{\mu_o \times 1.8 \times 4.0 \times l }{2 \pi \times 0.2} = \dfrac{\mu_o \times 4 \times I_3 \times l }{2 \pi \times 0.363}}[/tex]
Making I₃ the subject by equating both equations together, we have:
[tex]\mathbf{I_3 = \dfrac{1.8 \times 0.363}{0.2}}[/tex]
I₃ = 3.267 A and the current is pointing in the downward direction
Learn more about the magnitude and direction of forces here:
https://brainly.com/question/14879801?referrer=searchResults
Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with such a large total cross-sectional area that the blood velocity is much smaller: about 0.7 mm/s. Ignore viscosity in the capillaries. How long could your neck be so that blood still reaches your brain in m
The length of the neck that blood can still reach the brain depends on blood pressure, flow speed, and the cross-sectional area of capillaries. Using the given values, the approximate length of the neck is 9.8 meters.
Explanation:The length of your neck that blood can still reach your brain depends on the blood pressure, flow speed, and the cross-sectional area of the capillaries. The blood enters the capillaries in the brain with a much smaller velocity of about 0.7 mm/s due to the large total cross-sectional area. To determine the length of the neck, we need to consider the time it takes for blood to travel from the heart to the brain. Using the given velocity, we can calculate that the time it takes for blood to reach the brain is approximately 14 seconds. Assuming a constant velocity, the length of the neck can be calculated using the formula: length = velocity x time, which gives us a result of 9.8 meters.
Learn more about Blood flow to the brain here:https://brainly.com/question/32368169
#SPJ12
A water wave traveling in a straight line on a lake is described by the equation
y(x,t) = (3.30 cm) cos(0.400 cm?1x + 5.05 s?1t)
where y is the displacement perpendicular to the undisturbed surface of the lake.
(a) How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor?
What horizontal distance does the wave crest travel in that time?
b) What are the wave number and the number of waves per second that pass the fisherman?
(c) How fast does a wave crest travel past the fisherman?
What is the maximum speed of his cork floater as the wave causes it to bob up and down?
Final answer:
Calculations from the provided wave equation involve finding the period (T), horizontal wave travel, wave number and frequency, wave crest speed, and the maximum vertical displacement speed. These reveal wave motion characteristics including time intervals and speeds for complete wave patterns and individual points within a wave.
Explanation:
The student's question about a wave on a lake is one that involves understanding the characteristics and behavior of waves, which is a key concept in physics. Specifically, the student is asked to find the time for a complete wave pattern to pass a stationary observer, the horizontal distance traveled by a wave crest in that time, the wave number and frequency, the wave crest speed, and the maximum speed of a vertical displacement caused by the wave. Here's how we can calculate each of these:
Time for one complete wave pattern: The time taken for one complete wave to pass is the reciprocal of the frequency, known as the period (T). We use the angular frequency (5.05 s-1) provided in the equation y(x,t) = (3.30 cm) cos(0.400 cm-1x + 5.05 s-1t). We find T by taking 2π divided by the angular frequency.
Horizontal distance traveled: To find this, multiply the wave speed by the period (T).
Wave number and frequency: The wave number (k) is already provided as 0.400 cm-1, which can be converted to meters if needed. The frequency (f) is the angular frequency divided by 2π.
Wave crest speed: This can be found by dividing the angular frequency by the wave number.
Maximum speed of vertical displacement (cork floater): This is the amplitude times the angular frequency, which gives the maximum speed of any vertical movement due to the wave.
These calculations help understand the dynamics of wave motion and how waves interact with objects in their path like a fisherman's cork floater.
Fused quartz possesses an exceptionally low coefficient of linear expansion, 5.50 × 10 − 7 ( ∘ C ) − 1 . Suppose a bar of fused quartz of length 3.45 m at 20.0 ∘ C is heated to 235 ∘ C . By how much, in millimeters, would the bar expand?
To solve this problem we will use the concepts related to thermal expansion in a body for which the initial length, the coefficient of thermal expansion and the temperature change are related:
[tex]\Delta L = L0\alpha\Delta T[/tex]
Where,
[tex]\Delta L[/tex] = Change in Length
[tex]\alpha[/tex] = Coefficient of linear expansion
[tex]\Delta T[/tex] = Change in temperature
[tex]L_0[/tex] = Initial Length
Our values are:
[tex]L_0 = 3.45m[/tex]
[tex]\alpha = 5.5*10^{-7} \°C^{-1}[/tex]
[tex]\Delta T = 235-20 = 215\°C[/tex]
Replacing we have,
[tex]\Delta L = (3.49) (5.5*10^{-7}) [(215)[/tex]
[tex]\Delta L = 0.0004126m[/tex]
[tex]\Delta L = 0.4126mm[/tex]
Therefore the change in milimiters was 0.4126mm
Penguin #1 travels at speed v=0.9 c relative to penguin #2. At the instant that both are at the origin, they synchronize their clocks to both read 1:00 AM. Penguin #1 reports that her meter stick has length one meter. If #2 were to measure the length of the meter stick of #1, what would he observe its length to be?
Answer:
L = 0.44 [m]
Explanation:
Here we can use the Lorentz transformation related to length to solve it:
[tex]L=L_{0}\sqrt{1-\beta^{2}} [/tex]
Where:
L₀ is the length of the moving reference frame (penguin #1)
L is the length of the fixed reference frame (penguin #2)
β is the ratio between v and c
We know that v = 0.9c so we can find β.
[tex]\beta = \frac{0.9c}{c}=0.9[/tex]
[tex]L=1 [m]\sqrt {1-0.9^{2}} = 0.44 [m] [/tex]
Therefore, the length of the meter stick of #1 observed by #2 is 0.44 m.
I hope it helps you!
In fast-pitch softball, a pitcher swings her arm from straight overhead in a circle, releasing the 0.196 kg ball at the bottom of the swing. If a pitcher's arm has a length of 0.984 m and mass of 11.3 kg, and the ball goes from initially at rest to a tangential speed of 29.8 m/s before release, what average torque must the pitcher apply to the ball? Model the pitcher's arm as a uniform thin rod swung about one end, and make sure you don't forget the mass of the softball (but you can ignore its shape; i.e., treat it like a point mass). Hint: you will need to make use of the equations of circular motion for constant angular acceleration that we derived back in week 2.
Answer:
560.06714 Nm
Explanation:
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation = [tex]\pi[/tex] (Half rotation)
v = Velocity of bat = 29.8 m/s
M = Mass of bat = 11.3 kg
m = Mass of ball = 0.196 kg
R = Radius of swing = 0.984 m
[tex]\omega_f=\dfrac{v}{r}\\\Rightarrow \omega_f=\dfrac{29.8}{0.984}\\\Rightarrow \omega_f=30.28455\ rad/s[/tex]
From equation of rotatational motion
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{30.28455^2-0^2}{2\times \pi}\\\Rightarrow \alpha=145.96958\ rad/s^2[/tex]
Moment of inertia is given by
[tex]I=\dfrac{1}{3}MR^2+mR^2\\\Rightarrow I=\dfrac{1}{3}11.3\times 0.984^2+0.196\times 0.984^2\\\Rightarrow I=3.83687577\ kgm^2[/tex]
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow \tau=3.836875776\times 145.96958\\\Rightarrow \tau=560.06714\ Nm[/tex]
The torque the pitcher applies is 560.06714 Nm
The average torque this pitcher must apply to the softball is 560.64 Newton.
Given the following data:
Mass of ball = 0.196 kg.Mass of pitcher's arm = 11.3 kg.Length of pitcher's arm (radius) = 0.984 m.Initial speed = 0 m/s (since it starts from rest).Tangential speed = 29.8 m/s.To determine the average torque this pitcher must apply to the ball:
How to calculate the average torque.First of all, we would determine the final angular speed of the ball by using this formula:
[tex]\omega_f =\frac{v}{r} \\\\\omega_f =\frac{29.8}{0.984} \\\\\omega_f = 30.29\;rad/s[/tex]
Next, we would determine the constant angular acceleration by using the equation of circular motion:
[tex]\alpha =\frac{\omega^2_f - \omega^2_i}{2\theta} \\\\\alpha =\frac{30.29^2 - 0^2}{2\times \pi} \\\\\alpha =\frac{917.48}{2\times 3.142} \\\\\alpha =\frac{917.48}{6.284}\\\\\alpha = 146\;rad/s^2[/tex]
For the moment of inertia:
For a point mass, moment of inertia is given by this formula:
[tex]I =\frac{1}{3} Mr^2 + mr^2\\\\I =\frac{1}{3} \times 11.3 \times 0.984^2 + 0.196 \times 0.984^2\\\\I = 3.84 \;Kgm^2[/tex]
Now, we can determine the average torque:
[tex]Torque = I\alpha \\\\Torque =3.84 \times 146[/tex]
Torque = 560.64 Newton.
Read more on torque here: https://brainly.com/question/14839816
A machine gear consists of 0.10 kg of iron and 0.16 kg of copper.
How much total heat is generated in the part if its temperature increases by 35 C°? (Specific heats of iron and copper are 450 and 390 J/kg⋅°C, respectively.)
a. 910 Jb. 3 800 Jc. 4 000 Jd. 4 400 J
Answer:
option (c)
Explanation:
mass of iron = 0.10 kg
mass of copper = 0.16 kg
rise in temperature, ΔT = 35°C
specific heat of iron = 450 J/kg°C
specific heat of copper = 390 J/kg°C
Heat by iron (H1) = mass of iron x specific heat of iron x ΔT
H1 = 0.10 x 450 x 35 = 1575 J
Heat by copper (H2) = mass of copper x specific heat of copper x ΔT
H1 = 0.16 x 390 x 35 = 2184 J
Total heat H = H1 + H2
H = 1575 + 2184 = 3759 J
by rounding off
H = 4000 J
Answer:b
Explanation:
Given
mass of iron Gear [tex]m_{iron}=0.1 kg[/tex]
mass of copper [tex]m_{cu}=0.16 kg[/tex]
specific heat of iron [tex]c_{iron}=450 J/kg-^{\circ}C[/tex]
specific heat of copper [tex]c_{cu}=390 J/kg-^{\circ}C[/tex]
heat gain by iron gear [tex]=m_{iron}c_{iron}\Delta T[/tex]
[tex]Q_1=0.1\times 450\times 35=1575 J[/tex]
heat gain by iron gear [tex]=m_{cu}c_{cu}\Delta T[/tex]
[tex]Q_2=0.16\times 390\times 35=2184 J[/tex]
Total heat [tex]Q_{net}=Q_1+Q_2[/tex]
[tex]Q_{net}=1575+2184=3759 J\approx 3800 J[/tex]
4) An electron moving along the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, then the magnetic field must have a component
(A) along the +z-axis.
(B) along the -z-axis.
(C) along the -x-axis.
(D) along the +y-axis.
(E) along the -y-axis.
Answer:
B) along the -z-axis.
Explanation:
According to the Fleming's left hand rule we know that when a positive charge moves with velocity v perpendicular to the direction of magnetic field the it experiences a magnetic force perpendicular to the the plane in which the velocity vector and the magnetic field vector lies.As a result all the three vector are mutually perpendicular in such a case.Here as per given condition we have a negative charge moving along x-axis which experiences the deflection in -y direction so the magnetic field will be along the negative z axis i.e. into the screen.Answer:
option (b)
Explanation:
direction of velocity is along + X axis
direction of force is along - Y axis
So, by use of the
[tex]\overrightarrow{F}=q\left ( \overrightarrow{v} \times \overrightarrow{B}\right )[/tex]
Where, B be the magnetic field
As charge of electron is negative
so the direction of magnetic field is along negative z axis .
Choose True or False for all statements
1. When the Celsius temperature doubles, the Fahrenheit temperature doubles.
2. When heat is added to a system, the temperature must rise.
3. In a bimetallic strip of aluminum and brass which curls when heated, the aluminum is on the inside of the curve.
4. An iron plate has a hole cut in its center. When the plate is heated, the hole gets smaller.
5.When the pendulum of a grandfather clock is heated, the clock runs more slowly.
6.Cold objects do not radiate heat energy.
Answer:
Explanation:
1 . When the Celsius temperature doubles, the Fahrenheit temperature doubles - FALSE
A change of temperature from 300 F to 600 F changes celsius degree from 148 to 315 °C
2. When heat is added to a system, the temperature must rise.
- FALSE
When we add heat to a system of mixture of ice and water , temperature does not rise until whole of ice melts.
3. In a bimetallic strip of aluminum and brass which curls when heated, the aluminum is on the inside of the curve
- FALSE
Coefficient of linear thermal expansion of aluminium is more so it will be on the outer side.
4. An iron plate has a hole cut in its center. When the plate is heated, the hole gets smaller.
- FALSE
hole will be bigger. There will be photogenic expansion .
5.When the pendulum of a grandfather clock is heated, the clock runs more slowly.
TRUE
Time period is directly proportional to square root of length of pendulum.
when time period increases , clock becomes slow.
6.Cold objects do not radiate heat energy
- FALSE
Object radiates energy at all temperature .
6 .
The statements are (1)False, (2)False, (3)True, (4)False, (5)True, (6)False
Here are the answers to your statements:
False. When the Celsius temperature doubles, the Fahrenheit temperature does not double. The two scales are related by the formula [tex]F = \left(\frac{9}{5} \cdot C\right) + 32[/tex]False. When heat is added to a system, its temperature does not always necessarily rise; it might cause a phase change instead.True. In a bimetallic strip of aluminum and brass, the aluminum, having a higher coefficient of thermal expansion, will indeed be on the inside of the curve when heated.False. When an iron plate with a hole is heated, thermal expansion causes both the plate and the hole to enlarge.True. When the pendulum of a grandfather clock is heated, it expands, causing the clock to run more slowly due to the increase in the pendulum's length.False. Cold objects do still radiate heat energy, although they radiate less heat energy compared to hotter objects.A certain transverse wave is described by y(x,t)=Bcos[2π(xL−tτ)],where B = 6.40 mm , L = 26.0 cm , and τ = 3.90×10^−2 s .a. Determine the wave's amplitude.b. Determine the wave's wavelength.c. Determine the wave's frequency.d. Determine the wave's speed of propagation. e. Determine the wave's direction of propagation.
Answer
given,
[tex]y(x,t)=B cos[2\pi (\dfrac{x}{L} - \dfrac{t}{\tau})][/tex]
B = 6.40 mm , L = 26 cm , τ = 3.90 × 10⁻² s
general wave equation
y = A cos (k x - ωt)
where A is the amplitude of the
a) Amplitude of the given wave
B = 6.40 mm
b) Wavelength of the given wave
λ = L
λ = 26 cm
c) wave frequency
[tex]f = \dfrac{1}{\tau}[/tex]
[tex]f = \dfrac{1}{3.9 \times 10^{-2}}[/tex]
f = 25.64 Hz
d) speed of wave will be equal to
v = f λ
v = 25.64 x 0.26
v = 6.67 m/s
e) direction of propagation will be along +ve x direction because sign of k x and ωt is same as general equation.