A cart of mass 250 g is place on a frictionless horizontal air track. A spring having a spring constant of 9.5 N/m is attached between the cart and the left end o f the track. If the cart is displaced 4.5 cm from its equilibrium position, find: a) The period at which it oscillates.[1.0 s] b) Its maximum speed.[0.28 m/s] c) Its speed when it is located 2.0 cm from its equilibrium position.[0.25 m/s]

Answer:

1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2

2) .w is angular frequency. Answer 2

3) k  is wave number. Answer 1

Explanation:

The electromagnetic wave is given by

         [tex]E_{y}[/tex] = E₀ sin (kx –wt)

This is the equation of a traveling wave on the x axis with the elective field oscillating on the y axis

The terms represent E₀ the maximum amplitude of the electric field,

The wave vector

        k = 2π /λ

Angular velocity

       w = 2π f

To answer the questions let's use the previous definitions

1) Eo and Bo. They are maximum amplitudes. Answer 1 and 2

2) .w is angular frequency. Answer 2

3) k is wave number. Answer 1

Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, [tex]L=9.4\ mH=9.4\times 10^{-3}\ H[/tex]

Resistance, R = 150 ohms

Capacitance, [tex]C=1.9\ \mu F=1.9\times 10^{-6}\ C[/tex]

At resonance, the capacitive reactance is equal to the inductive reactance such that,

[tex]X_C=X_L[/tex]    

[tex]2\pi f_o L=\dfrac{1}{2\pi f_oC}[/tex]

f is the resonant frequency of this circuit  

[tex]f_o=\dfrac{1}{2\pi \sqrt{LC}}[/tex]

[tex]f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}[/tex]

[tex]f_o=1190.91\ Hz[/tex]

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

Answer:

0.247

5.25×10⁻¹³ J

Explanation:

Part 1/2

Elastic collision means both momentum and energy are conserved.

Momentum before = momentum after

m v = m v₁ + 14.1m v₂

v = v₁ + 14.1 v₂

Energy before = energy after

½ m v² = ½ m v₁² + ½ (14.1m) v₂²

v² = v₁² + 14.1 v₂²

We want to find the fraction of the neutron's kinetic energy is transferred to the atomic nucleus.

KE/KE = (½ (14.1m) v₂²) / (½ m v²)

KE/KE = 14.1 v₂² / v²

KE/KE = 14.1 (v₂ / v)²

We need to find the ratio v₂ / v.  Solve for v₁ in the momentum equation and substitute into the energy equation.

v₁ = v − 14.1 v₂

v² = (v − 14.1 v₂)² + 14.1 v₂²

v² = v² − 28.2 v v₂ + 198.81 v₂² + 14.1 v₂²

0 = -28.2 v v₂ + 212.91 v₂²

0 = -28.2 v + 212.91 v₂

28.2 v = 212.91 v₂

v₂ / v = 28.2 / 212.91

v₂ / v = 0.132

Therefore, the fraction of the kinetic energy transferred is:

KE/KE = 14.1 (0.132)²

KE/KE = 0.247

Part 2/2

If a fraction of 0.247 of the initial kinetic energy is transferred to the atomic nucleus, the remaining 0.753 fraction must be in the neutron.

Therefore, the final kinetic energy is:

KE = 0.753 (6.98×10⁻¹³ J)

KE = 5.25×10⁻¹³ J

When a neutron collides elastically with the nucleus of an atom, a fraction of its kinetic energy is transferred to the nucleus. The fraction of kinetic energy transferred can be calculated using the principle of conservation of momentum and kinetic energy. For the given scenario, the fraction is 0.8636. To find the final kinetic energy of the neutron, multiply the fraction of kinetic energy transferred by the initial kinetic energy of the neutron.

When a neutron in a reactor undergoes an elastic head-on collision with the nucleus of an atom initially at rest, kinetic energy is transferred from the neutron to the atomic nucleus. The fraction of the neutron's kinetic energy transferred to the nucleus can be calculated using the principle of conservation of momentum and kinetic energy. Since the mass of the atomic nucleus is about 14.1 times the mass of the neutron, the fraction of kinetic energy transferred can be calculated as:

Fraction of kinetic energy transferred = (14.1 - 1) / (14.1 + 1) = 0.8636

For PART 2/2, to find the final kinetic energy of the neutron, we can multiply the fraction of kinetic energy transferred to the nucleus by the initial kinetic energy of the neutron:

Final kinetic energy = Fraction of kinetic energy transferred x Initial kinetic energy = 0.8636 x 6.98 × 10-13 J

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Answer:

75.71 m/s

Explanation:

From equation of motion, acceleration is given by

[tex]a=\frac {v-u}{t}[/tex]where v is the final velocity, u is the initial velocity and t is time taken.

Making v the subject of the above formula

v=at+u

Substituting 6.7 s for time, t and 11.3 for a and taking u as zero since it starts from rest

v=11.3*6.7=75.71 m/s

Answer:

a. 4 J/s

Explanation:

Fourier's law states for the case in which there is stationary heat flow in only one direction, that is, linearly, the heat transmitted per unit of time is proportional to the temperature difference:

[tex]\frac{Q}{t}\propto \Delta T[/tex]

When the temperature at one end is 100°C and at the other is 20°C, we have:

[tex]\Delta T_1=100^\circ C-20^\circ C\\\Delta T_1=80^\circ C[/tex]

If the temperature of the hotter end is [tex]80^\circ C[/tex], we have:

[tex]\Delta T_2=100^\circ C-80^\circ C\\\Delta T_2=20^\circ C[/tex]

So:

[tex]\Delta T_1=4\Delta T_2\\\Delta T_2=\frac{\Delta T_1}{4}[/tex]

Finally, we calculate the rate of heat transfer:

[tex]\frac{Q_2}{t_2}=\frac{\frac{Q_1}{t_1}}{4}\\\\\frac{Q_2}{t_2}=\frac{16\frac{J}{s}}{4}\\\frac{Q_2}{t_2}=4\frac{J}{s}[/tex]

Answer:

Ec=53.7×10⁹N/m² =53.7Gpa

Explanation:

To calculate the modulus of elasticity in the longitudinal direction.  This is possible realizing Ec=σ/ε where σ=(Fm+Ff)/Ac

[tex]Ec=Sigma/E\\Ec=\frac{(Fm+Ff)/E}{Ac}\\ Ec=\frac{1802+74,000}{(1.25*10^{-3})(1130)(1/1000)^{2}  }\\ Ec=53.7*10^{9}N/m^{2}\\or\\Ec=53.7GPa[/tex]

Final answer:

The modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.

Explanation:

To find the modulus of elasticity of the composite material in the longitudinal direction, we can use the formula:

E = (stress sustained by the fiber phase)/(longitudinal strain)

Given that the stress sustained by the fiber phase is 156 MPa and the total longitudinal strain is 1.25 x 10^-3, we can plug in these values to calculate the modulus of elasticity:

E = 156 MPa / (1.25 x 10^-3) = 124,800 MPa

Therefore, the modulus of elasticity of the composite material in the longitudinal direction is 124,800 MPa.

Answer:

2258.4 m

Explanation:

Distance covered is a product of speed and time hence

s=vt where s is the displacement/distance covered, v is the speed and t is the time taken

s=24*94.1=2258.4 m

Therefore, the distance covered is 2258.4 m

Answer:

3

Explanation:

During a crash 3 types of collisions can occur.

There are usually three types of collisions involved in a motor vehicle crash.

The first type of collision is the vehicle collision, which involves the physical impact between two or more vehicles. When vehicles collide, their structures deform, and the forces involved can cause severe damage to the involved vehicles. The severity of this collision depends on factors like the speed, mass, and angle of impact.

The second type of collision is the human collision, which occurs inside the vehicle. During an accident, passengers inside the vehicle can collide with each other or with interior components, such as the dashboard, steering wheel, or windows. These collisions can result in injuries like whiplash, head injuries, or broken bones.

The third type of collision is the internal collision, which involves the organs and tissues within the human body. When a collision occurs, the human body is subjected to rapid deceleration, causing organs to collide with each other or with the skeletal structure. These internal collisions can lead to internal bleeding, organ damage, and other life-threatening injuries.

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Answer:

a. Ek = 62.5 x 10⁻³ J

b. Ek = 31.25 x 10 ⁻³ J

Explanation:

E = [ Us + (M * v²) / 2 ] + [ (I * ω² ) / 2 ]

T = 2π * √ 3M / 2 k , K = 3.0 N / m , d = 0.25m

a.

Ek = ¹/₂ * m * v² = ¹/₃ * k * d²

Ek = ¹/₃ * k * d²  = ¹/₃ * 3.0 N / m * 0.25²m

Ek = 62.5 x 10⁻³ J

b.

Ek = ¹/₂ * I * ω² = ¹/₄ * M * v²

Ek = ¹/₆ * k * Xm² = ¹/₆ * 3.0 N / m * 0.25²m

Ek = 31.25 x 10 ⁻³ J

c.

d Emech / dt = d / dt * [ 3 m * v² / 4 + k * x² / 2 ]

acm = - ( 2 k / 3M)

ω = √ 2k / 3m     ⇒  T = 2π * √ K / m

Answer:

a) 0.0674s

b) 0.3880s

Explanation:

Based on the speed of sound in each medium ([tex]s_{air}=339m/s, s_{water}=1480m/s, s_{metal}=5060m/s[/tex]) the sound will first arrive via the metal handrail, then the water and lastly the air.

[tex]t=\frac{d}{s}[/tex] , where 't' is time, 'd' is distance and 's' is speed

[tex]t_{metal}=\frac{d}{s_{metal}}[/tex]

[tex]t_{metal}=\frac{141}{5060}[/tex]

[tex]t_{metal}=0.0279s[/tex]

[tex]t_{water}=\frac{d}{s_{water}}[/tex]

[tex]t_{water}=\frac{141}{1480}[/tex]

[tex]t_{water}=0.0953s[/tex]

[tex]t_{air}=\frac{d}{s_{air}}[/tex]

[tex]t_{air}=\frac{141}{339}[/tex]

[tex]t_{air}=0.4159s[/tex]

a) Time difference between the first and second sound

[tex]t_{water}-t_{metal}[/tex]

[tex]0.0953-0.0279[/tex]

[tex]0.0674s[/tex]

b) Time difference between the first and third sound

[tex]t_{air}-t_{metal}[/tex]

[tex]0.4159-0.0279[/tex]

[tex]0.3880s[/tex]

The second sound arrives 0.321 seconds after the first, and the third sound arrives 0.388 seconds after the first sound, demonstrating how sound speed varies in different media.

The question involves calculating the time differences for sound to travel through different media (air, water, and a metal handrail) over the same distance. To find out how much later one sound arrives compared to another, we use the formula time = distance / speed. The distances for all three media are the same, 141 m, but the speeds vary: 339 m/s for air, 1480 m/s for water, and 5060 m/s for the metal handrail.

Air: time = 141 m / 339 m/s = 0.416 s

Water: time = 141 m / 1480 m/s = 0.095 s

Metal handrail: time = 141 m / 5060 m/s = 0.028 s

(a) The second sound (through water) arrives 0.416 s - 0.095 s = 0.321 s after the first sound (through air).

(b) The third sound (through the metal handrail) arrives 0.416 s - 0.028 s = 0.388 s after the first sound (through air).

The speed of the ship remains the same when vacationers run from the front to the back. This effect is explained by the conservation of momentum, which states that the total momentum of an isolated system, in this case the ship and the vacationers, remains constant. Thus, the internal movement of the vacationers does not affect the speed of the ship.

This question is related to the principle of conservation of momentum in physics. According to this principle, if a system is isolated (no external forces acting on it), the total momentum remains constant. So, the spectators' running from the bow to the stern won't affect the speed of the ship. Therefore, the answer is 2. The speed of the ship is unchanged from what it was before they started running. It is important to note that the center of mass of the system (ship and vacationers) remains the same before and after the vacationers run from one point to another inside the ship. This is because their movement is internal to the system and has no effect on the system's center of mass or the ship's speed.

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The speed of the cruise ship remains unchanged even if the vacationers move from the bow to the stern. This is due conservation of linear momentum and Newton's third law. Even though the ship's center of mass is slightly affected, the total speed of the ship remains the same.

Physics principles, particularly those related to Newton's third law and the conservation of linear momentum, apply in this scenario. As the vacationers from the bow run towards the stern, their forward motion will push the ship slightly backward due to Newton's third law, which states that every action has an equal and opposite reaction. However, the overall speed (or velocity) of the cruise ship relative to the water or the Earth remains unchanged, option 2, because the total linear momentum of the cruise ship system (which includes the ship and the passengers) is conserved.

It's important to note that the shift in passengers’ positions does slightly change the center of mass of the entire system, but it does not alter the fact that the total speed of the ship is conserved if no external force is applied.

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The work done by vector field F on a particle moving along curve C can be computed using the line integral ∫F.dr. To compute the line integral, we need to parameterize curve C and evaluate the dot product of the vector field and the parameterized curve.

To compute the work done by a vector field F on a particle moving along a curve C, we can use the line integral. The line integral of a vector field F along a curve C can be computed using the formula: ∫F.dr. In this case, F(x, y, z) = xi + yj + zk and C is given by r(t) = (1 + 2sin(t))i + (1 + 3sin(2t))j + (1 + sin(3t))k. We need to parameterize C to compute the line integral. Let's rewrite r(t) as:

r(t) = i + 2sin(t)i + j + 3sin(2t)j + k + sin(3t)k

We can then calculate the line integral using the given parameterization. Substituting r(t) into F(x, y, z), we get:

F(r(t)) = (1 + 2sin(t))i + (1 + 3sin(2t))j + (1 + sin(3t))k

Now, we can compute the line integral by evaluating ∫F(r(t)).dr over the given interval 0 ≤ t ≤ π/2. This involves evaluating the dot product of F(r(t)) and r'(t). The work done by F on the particle moving along C is the value of the line integral.

Final answer:

To compute the work done by a vector field on a particle moving along a curve, we can use the line integral formula. In this case, the curve C is given by r(t) = (1+2sin(t))i + (1+3sin(2t))j + (1+sin(3t))k, where 0 ≤ t ≤ π/2. The vector field F(x, y, z) = xi + yj + zk. The work done is equal to the line integral of F along C.

Explanation:

To compute the work done by a vector field on a particle moving along a curve, we can use the line integral formula:

Work = ∫C F · dr

In this case, the curve C is given by r(t) = (1+2sin(t))i + (1+3sin(2t))j + (1+sin(3t))k, where 0 ≤ t ≤ π/2. The vector field F(x, y, z) = xi + yj + zk.

The work done is equal to the line integral of F along C, so:

Work = ∫0^π/2 F(r(t)) · r'(t) dt

Now, we can substitute the given expressions for F and r(t) and evaluate the integral to find the work done.

Answer:

[tex]1.91738\times 10^{-20}\ kgm/s[/tex]

[tex]1.05456\times 10^{-24}\ kgm/s[/tex]

Explanation:

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

[tex]\Delta x[/tex] = Uncertainty in the position

[tex]\Delta p[/tex] = Uncertainty in the momentum

From the uncertainty principle

[tex]\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 5.5\times 10^{-15}}\\\Rightarrow \Delta p=1.91738\times 10^{-20}\ kgm/s[/tex]

The minimum uncertainty in the momentum of the proton is [tex]1.91738\times 10^{-20}\ kgm/s[/tex]

[tex]\Delta x\Delta p=\dfrac{h}{2\pi}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{h}{2\pi \Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{2\pi\times 1\times 10^{-10}}\\\Rightarrow \Delta p=1.05456\times 10^{-24}\ kgm/s[/tex]

The minimum uncertainty in the momentum of the electron is [tex]1.05456\times 10^{-24}\ kgm/s[/tex]

Final answer:

The minimum uncertainty in the momentum of a proton confined to the nucleus of an atom is 9.6 × 10^-12 kg m/s. The minimum uncertainty in the momentum of an electron confined in an atom is 5.3 × 10^-24 kg m/s.

Explanation:

The Heisenberg Uncertainty Principle states that there is a limit to the precision with which we can know both the position and momentum of a particle. The minimum uncertainty in the proton's momentum is given by the formula: Dpmin = ħ/2Ax, where Ax is the uncertainty in the position of the proton. In this case, Ax is given as the diameter of the nucleus, so we have:

Dpmin = (1.055 × 10^(-34) kg m^2/s) / (2(5.5 × 10^(-15) m)) = 9.6 × 10^(-12) kg m/s.

Similarly, for the electron:

Dpmin = (1.055 × 10^(-34) kg m^2/s) / (2(1 × 10^(-10) m)) = 5.3 × 10^(-24) kg m/s.

The period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].

Step 1

The period T of a simple pendulum is given by the formula:

[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]

where:

- T is the period of the pendulum,

- L is the length of the pendulum, and

- g is the acceleration due to gravity.

To find the period [tex]\( T_{\text{Mars}} \)[/tex] on the surface of Mars, we can use the formula with the acceleration due to gravity on Mars, [tex]\( g_{\text{Mars}} = 3.71 \, \text{m/s}^2 \)[/tex], and the same length L as on Earth.

We have the period [tex]\( T_{\text{Earth}} = 2.00 \, \text{s} \)[/tex] on Earth, and we need to find [tex]\( T_{\text{Mars}} \).[/tex]

Step 2

Using the formula, we get:

[tex]\[ T_{\text{Earth}} = 2\pi \sqrt{\frac{L}{g_{\text{Earth}}}} \][/tex]

Solving for L, we find:

[tex]\[ L = \left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}} \][/tex]

Now, using this value of L, we can find [tex]\( T_{\text{Mars}} \)[/tex] using the acceleration due to gravity on Mars:

[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{L}{g_{\text{Mars}}}} \][/tex]

Substituting the known values, we get:

[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{\left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}}}{g_{\text{Mars}}}} \][/tex]

Step 3

Simplifying:

[tex]\[ T_{\text{Mars}} = T_{\text{Earth}} \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Mars}}}} \]Now, let's calculate \( T_{\text{Mars}} \):\[ T_{\text{Mars}} = 2.00 \times \sqrt{\frac{9.81}{3.71}} \]\[ T_{\text{Mars}} = 2.00 \times \sqrt{2.643} \]\[ T_{\text{Mars}} \approx 2.00 \times 1.626 \]\[ T_{\text{Mars}} \approx 3.25 \, \text{s} \][/tex]

So, the period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].

Answer:

77362.56 J

163730.28571 J

Explanation:

A = Area = 25 mm²

l = Length = 300 mm

K = Constant = [tex]3.33\times 10^{-6}[/tex]

[tex]\eta[/tex] = Heat transfer factor = 0.75

[tex]f_m[/tex] = Melting factor = 0.63

T = Melting point of low carbon steel = 1760 K

Volume of the fillet would be

[tex]V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3[/tex]

The unit energy for melting is given by

[tex]U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3[/tex]

Heat would be

[tex]Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J[/tex]

Heat required to weld is 77362.56 J

Amount of heat generation is given by

[tex]Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J[/tex]

The heat generated at the welding source is 163730.28571 J

Without the specific heat capacity and melting point of low carbon steel, the exact heat required for the weld cannot be calculated. However, given the heat transfer factor and melting factor, the heat generated at the welding source is 163,700 Joules according to the student's provided answer.

The quantity of heat required for welding low carbon steel with a fillet weld having a cross-sectional area of 25.0 mm2 and length of 300 mm depends on the specific heat capacity of the steel and the temperature change required to melt it. However, the question does not provide specific values for the specific heat capacity or the melting point of low carbon steel, which are essential to calculate the heat quantity. Normally, such calculations would also require knowledge of the latent heat of fusion for the steel.

Given the lack of necessary details to calculate the quantity of heat, we can address the part (b) of the question which relates to the heat generated at the welding source. The heat generated at the source can be calculated by dividing the actual heat needed to make the weld (which is given by the student as an unknown, hence represented by the question mark '?') by the product of the heat transfer factor and the melting factor, which are 0.75 and 0.63 respectively.

If the heat required to perform the weld ('?') is found, then the heat generated at the source can be calculated as follows: Heat at source = Heat required / (Heat transfer factor × Melting factor). According to the answer provided by the student, the heat at the source is 163,700 Joules.

Answer:

What material would you use: I would use a beaker, water, heated metal

What would you measure: Measure the changes in temperature before and after pouring the metal.

What results would you expect: Whether Thermal Equilibrium has occurred or not.

What if the results were different; what would that indicate : The Beaker might absorb some of the heat causing an error in the reading.

Explanation:

step 1: Take a well insulated beaker and pour water of known mass in it.

step 2: Record its initial temperature.

step 3:  Place heated metal into the beaker and make sure that the beaker is tightly packed so that no heat escapes from the beaker.

step 4: Record temperature of the beaker at different intervals and after temperature has become constant ( No heat is being gained) , note the final temperature

In end we will check whether thermal equilibrium is established or not.

heat lost by the metal = heat gained by water + heat gained by the beaker    

Possible Error:

There might be disparities in the values acquired as the beaker will absorb some heat.

Answer:

(a) Velocity at time t will be  [tex]v(t)=t^2+4t-32[/tex]

(B) Distance will be -48 m

Explanation:

We have given [tex]a(t)=2t+4[/tex]

And [tex]v(0)=-32[/tex]

(a) We know that [tex]v(t)=\int a(t)dt[/tex]

So [tex]v(t)=\int (2t+4)dt[/tex]

[tex]v(t)=t^2+4t+c[/tex]

As [tex]v(0)=-32[/tex]

So [tex]-32=0^2+4\times 0+c[/tex]

c = -32

So [tex]v(t)=t^2+4t-32[/tex]

(b) We have to find the distance traveled

So [tex]s(t)=\int_{0}^{6}v(t)dt[/tex]

[tex]s(t)=\int_{0}^{6}(t^2+4t-32)dt[/tex]

[tex]s(t)=\int_{0}^{6}(\frac{t^3}{3}+2t^2-32t)[/tex]

[tex]s=(\frac{6^3}{3}+2\times 6^2-32\times 6)-0=72+72-192=-48m[/tex]

Final answer:

To find the velocity at time t, integrate the acceleration function and add the initial velocity. To find the distance traveled, integrate the velocity function over the given time interval.

Explanation:

To find the velocity at time t, we can integrate the acceleration function over the given time interval and add the initial velocity. The integral of 2t + 4 is t^2 + 4t, so the velocity function is v(t) = t^2 + 4t - 32.

To find the distance traveled, we can integrate the velocity function over the given time interval. The integral of t^2 + 4t - 32 is (1/3)t^3 + 2t^2 - 32t. Evaluating this integral from 0 to 6 gives us the distance traveled during the given time interval.

Answer:

Normal force, N = 60000 N                      

Explanation:

It is given that,

Mass of the automobile, m = 2000 kg

Radius of the curved road, r = 20 m

Speed of the automobile, v = 20 m/s

Let N is the normal and F is the net force acting on the automobile or the centripetal force. It is given by :

[tex]N-mg=\dfrac{mv^2}{r}[/tex]

[tex]N=\dfrac{mv^2}{r}+mg[/tex]    

[tex]N=m(\dfrac{v^2}{r}+g)[/tex]                

[tex]N=2000\times (\dfrac{(20)^2}{20}+10)[/tex]                            

N = 60000 N  

So, the normal force exerted by the road on the system is 60000 Newton. Hence, this is the required solution.

Final answer:

The normal force exerted by the road on the system (car and its occupants) when moving through a curved dip at a constant speed, calculated considering both gravitational and centripetal forces, is 60,000 N.

Explanation:

To calculate the normal force exerted by the road on the system (car and its occupants), we first note that when an object is in circular motion, the net force acting on the object is directed towards the center of the circle. In this case, the net force is the centripetal force required to keep the automobile in circular motion, which can be calculated using the formula Fc = m*v2/R where m is the mass of the automobile, v is its velocity, and R is the radius of the curve.

For the given values (m = 2000 kg, v = 20 m/s, and R = 20 m), the centripetal force is calculated as Fc = 2000 kg * (20 m/s)2 / 20 m = 2000 kg * 400 m2/s2 / 20 m = 40,000 N. This force is provided by the component of the normal force that acts towards the center of the circular path. Additionally, the normal force must counteract the gravitational force acting on the automobile (Fg = m*g), which is 2000 kg * 10 m/s2 = 20,000 N.

However, in the scenario of a car moving through a curved dip, the normal force also provides the centripetal force. The total normal force exerted by the road must therefore support the weight of the car and provide the centripetal force needed for circular motion. Thus, the total normal force is N = Fg + Fc = 20,000 N + 40,000 N = 60,000 N.

Final answer:

The sliding distance of the box can be calculated using the principles of conservation of energy and the work-energy theorem, considering the initial kinetic energy and the work done against the force of kinetic friction.

Explanation:

Calculating the Sliding Distance of a Box on an Inclined Plane

To determine how far the box slides, we can use the principles of conservation of energy and the work-energy theorem. The initial kinetic energy of the box is transformed into work done against friction. The work done by friction is equal to the force of friction times the distance the box slides. We start with calculating the force of kinetic friction, which is μ_k (coefficient of kinetic friction) times the normal force. The normal force is the component of the box's weight perpendicular to the inclined plane, calculated as m*g*cos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline.

With given values: m = 17.6 kg, μ_k = 0.48, θ = 19°, βi = 2.25 m/s, and g = 9.8 m/s², we can calculate the force of kinetic friction (ƒ_k).

The component of gravity along the incline is m*g*sin(θ), and we know that the box stops when its initial kinetic energy is equal to the work done by friction. So, from the equation:

Kinetic Energy_initial = Work_friction,

½*m*βi^2 = ƒ_k * distance,

½*17.6 kg*(2.25 m/s)^2 = (0.48*17.6 kg*9.8 m/s²*cos(19°)) * distance,

We can then solve for the distance the box slides. After calculation, we obtain the sliding distance x.

The correct step for obtaining a common denominator in the expression 'a = a1 + F/m' is to multiply each term by m/m, which results in '(m/m * a1/1) + (m/m * F/m)'. This leads to the acceleration, a, equalling 4.71 m/s2 when substituting the given values into the expression.

The student is trying to solve for the acceleration (a) of an object using the equation a = a1 + F/m, where a1 is given as 3.00 meters/second2, F as 12.0 kilogram.meter/second², and m as 7.00 kilograms. To obtain a common denominator for the two terms a1 and F/m, we look for an expression that allows us to combine these two fractions.

The correct step for obtaining a common denominator is option d. This is written as:

(m/m * a1/1) + (m/m * F/m)

This is because multiplying by m/m is equivalent to multiplying by 1, which does not change the value of the expression. For the term a1, since there is no denominator, multiplying by m/m effectively gives it a common denominator with F/m. The calculation becomes:

a = (m * a1 + F) / m

Substituting the given values:

a = (7.00 kg * 3.00 m/s2 + 12.0 kg*m/s2) / 7.00 kg

= (21.00 + 12.00) kg*m/s2 / 7.00 kg

= 33.00 kg*m/s2 / 7.00 kg

= 4.71 m/s2

Thus, the newton's second law, a = F/m, can be used to calculate the acceleration of the object.

Answer:

B. Balanced Forces

Explanation:

The net force is defined as the sum of all the forces acting on an object. Therefore, if the forces are balanced, they will counteract each other, causing the net force to be zero, then the object will continue at rest or moving with constant velocity.

The term below best describes the forces on an object with a a net force of zero B. Balanced Force

Balanced Forces refer to the situation where the forces acting on an object are equal in magnitude and opposite in direction, resulting in a net force of zero. When balanced forces act on an object, they cancel each other out, leading to no change in the object's state of motion.

This concept is tied to Newton's First Law of Motion, which states that an object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless acted upon by an unbalanced force.

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Answer:

F= 9.45 N

Explanation:

If the mass is fastened to an unstrained horizontal spring, this means that at this position, the spring doesn't exert any force, because it keeps his equilibrium length.

If then the block is given a displacement of +0.113m, this means that the spring has been stretched in the same length.

According to Hooke's Law, the spring exerts a restoring force (trying to return to his equilibrium state) that opposes to the displacement, and which is proportional (in magnitude) to it, being the proportionality constant, a quantity called spring constant, which depends on the type of spring.

We can write the Hooke's Law as follows:

F = - k * Δx

Just before the block is released, we can get the value of F as follows:

⇒ F = 83.6 N/m* 0.113 m = 9.45 N (in magnitude)

Answer:

b. 84 minutes

Explanation:

[tex]a_c=g[/tex] = Centripetal acceleration = 9.81 m/s²

r = Radius of Earth = 6378.1 km

v = Velocity

Centripetal acceleration is given by

[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s[/tex]

Time period is given by

[tex]T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes[/tex]

The time period of Earth’s rotation would be 84.43835 minutes

The new period of the Earth’s rotation is mathematically given as

T=84.43835 min

Question Parameter(s):

The Earth’s radius is 6378.1 kilometers.

g= (9.81 m/s2 ),

Generally, the equation for the   is mathematically given as
[tex]a_c=\dfrac{v^2}{r}[/tex]

Therefore

[tex]v=\sqrt{a_cr}\\\\v=\sqrt{9.81*6378100}[/tex]

v=7910.06706 m/s

In conclusion

[tex]T=\dfrac{2\pi r}{v60}[/tex]

Hence

[tex]T=\dfrac{2\pi 6378.1*10^3}{7910.06706*60}[/tex]

T=84.43835 min

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Answer:

option C.

Explanation:

The correct answer is option C.

Vernier Calipers is a precision instrument that is used to measure the internal or outer dimensions or the depth of the material.

Vernier Caliper consists of two jaws, the main scale, and the vernier scale.

Graduation is present on both the main scale and vernier.

To measure the width of any material it should be placed in between the jaws and reading is taken with the help of the main scale and vernier scale.

The thickness of the candle can be measured using vernier calipers precisely.

Answer:

Consider the following calculations

Explanation:

a )  velocity of the glass is v = distance / time

= 5 X 10-3 / 7 X 365.24 X 24 X 60 X 60

v = 2.263 X 10-11 m/sec

the speed of the light in vaccum is C

C = 3 X 108 m/sec

n = C / v

n = 3 X 108 / 2.263 X 10-11

n = 1.32567 X 1019

b )  given is 40 km/hr

= 40 X 103 / 60 X 60

= 11.11 m/sec

n = C / v

n = 3 X 108 / 11.11

n = 27002700.27

The index of refraction for the 5.00 mm thick slow glass taking light 7.00 years to pass through is approximately 1.33 x 10^19. For a Bose-Einstein condensate where light travels at 40.0 km/hr, the effective index of refraction is about 2.70 x 10^7.

In Bob Shaw's short story 'The Light of Other Days', a fictional material called slow glass is described, which delays the passage of light. To calculate the index of refraction of a 5.00 mm thick piece of slow glass where light takes 7.00 years to pass through, we can use the formula n = c/v, where c is the speed of light in a vacuum (3.00 x 108 m/s), and v is the speed of light through the material.

To find v, we can calculate the total distance light travels in 7.00 years and divide it by the time it takes to travel through the slow glass. Since the slow glass is 5.00 mm thick, which is equivalent to 5.00 x 10-3 m, and one year is 365.24 days, we calculate the speed as follows:

v = distance/time = 5.00 x 10-3 m / (7.00 years x 365.24 days/year x 24 hours/day x 3600 seconds/hour) = 5.00 x 10-3 m / 220,937,280 seconds ≈ 2.263 x 10-11 m/s.

Then, the index of refraction, n, can be calculated as n = c/v ≈ 3.00 x 108 m/s / 2.263 x 10-11 m/s ≈ 1.33 x 1019.

For Lene Hau's experiment using a Bose-Einstein condensate with a light speed of 40.0 km/hr, the index of refraction can also be calculated using n = c/v. Converting 40.0 km/hr to m/s:

v = 40.0 km/hr x (1000 m/km) / (3600 s/hr) = 11.1 m/s.

Using this value for v, we calculate n as n = c/v ≈ 3.00 x 108 m/s / 11.1 m/s ≈ 2.70 x 107.

Answer

given,

force = 94 lb

weight of crate = 220 lb

Assuming the static friction be equal = 0.47

                       kinetic friction = 0.36

Maximum force applied to move the object is when object is just start to move.

F = μ N

F = 0.47 x 220

F = 103.4 lb

As the frictional force is more than applied then the object will not move.

so, the friction force will be equal to the force applied on the object that is equal to 94 lb.

hence, the direction of force will left.

Answer:

Explanation:

1 )  tire of radius 0.381 m rotating at 12.2 rpm

12.2 rpm = 12.2 /60 rps

n = .20333 rps

angular speed

= 2πn

= 2 x 3.14 x .20333

= 1.277 rad / s

2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s

angular speed = .456 rad/s

3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second

18.7° per second = (18.7 / 180) x 3.14 rad/s

= .326  rad/s

4 )

a rock on a string being swung in a circle of radius 0.587 m with

a centripetal acceleration of 4.53 m/s2

centripetal acceleration = ω²R

ω is angular velocity and R is radius

4.53 = ω² x .587

ω  = 2.78 rad / s

5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s

The radius of the circle in which corner is moving

= .123 x √2

=.174 m

angular velocity = linear velocity / radius

.287 / .174

1.649 rad / s

The perfect order is

4 ) > 5> 1 >2>3.

Explanation:

To arrange the listed objects according to their angular speeds from largest to smallest, we must first calculate the angular speed ω (in radians per second) for each object, as this unit provides a common ground for comparison. The angular speed can be found using the formula ω = v/r where v is the linear velocity and r is the radius.

Arranging these angular speeds from largest to smallest:

This event is a result of interference, a physics phenomenon where waves combine to create a new wave. Changing the frequency altered the phase difference between the waves at various places in the room, leading to some students now hearing a tone because they are at points of constructive interference.

The correct answer is c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone. This situation is described by the phenomenon of interference. Interference occurs when two waves combine to form a resultant wave. When two identical sound waves from the speakers meet at a point in space, they can either constructively or destructively interfere depending on the phase difference.

If the phase difference is such that the waves reinforce each other, it results in a loud sound (constructive interference). However, if the phase difference is such that one wave cancels the other, no sound is heard (destructive interference). By doubling the frequency, the professor effectively changes the phase difference between the waves at the various points in the room. Therefore, some students' locations might now be at points of constructive interference, allowing them to hear the sound.

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The correct answer is option a. When the oscillator's frequency is doubled, the interference pattern shifts, causing some students who originally heard a loud tone to hear nothing, while others still hear a loud tone.

The student asked what happens when the professor adjusts the oscillator to produce sound waves of twice the original frequency. The correct answer is a)  Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.

The phenomenon described in the question is due to the interference of sound waves. Interference occurs when sound waves from the two speakers overlap, creating areas of constructive interference (where waves are in phase and amplify the sound) and destructive interference (where waves are out of phase and cancel each other out).

When the frequency is changed, the interference pattern shifts, causing some students who previously experienced a loud sound to now be in a zone of destructive interference, and vice versa. This adjustment results in a new distribution of loud and quiet spots within the classroom.

Answer:

1.8 cm

Explanation:

[tex]m[/tex] = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

[tex]q[/tex] = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

[tex]\Delta V[/tex] =Potential difference through which the ion is accelerated = 215 V

[tex]v[/tex] = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

[tex](0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}[/tex]

[tex]r[/tex] = Radius of the path followed by ion

[tex]B[/tex] = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

[tex]qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm[/tex]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

[tex]M = \frac{Nl}{f_ef_0}[/tex]

Where

N = Near point

l = distance between the object lens and eye lens

[tex]f_0[/tex]= Focal length

[tex]f_e[/tex]= Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

[tex]M = \frac{25*10}{3*5}[/tex]

[tex]M = 16.67\approx 17\\[/tex]

Therefore the correct answer is C.