Explanation:
Diameter of Sun = 100 x Diameter of Earth.
We are creating a model where Earth is same as golf ball,
Diameter of Earth in model = Diameter of golf ball
Diameter of Earth in model = 1.68 inches
Diameter of Sun in model = 100 x Diameter of Earth in model
Diameter of Sun in model = 100 x 1.68
Diameter of Sun in model = 168 inches = 14 feet
Diameter of Sun in model = 168 inches = 14 feet = 4.27 m
A 555 g ball strikes a wall at 14.9 m/s and rebounds at 12.1 m/s. The ball is in contact with the wall for 0.044 s. What is the magnitude of the average force acting on the ball during the collision?
Final answer:
The magnitude of the average force acting on the 555 g ball during the collision is approximately 347.39 N. This is calculated using the impulse-momentum theorem, as the product of the average force and the contact time equals the change in momentum.
Explanation:
The student's question pertains to the concept of impulse and the average force exerted during a collision, which are foundational topics in dynamics, a branch of Physics. To find the magnitude of the average force acting on the ball during the collision, we can use the impulse-momentum theorem. The impulse exerted on an object is equal to the change in momentum of the object, and it can also be calculated as the average force times the time period of the force's application.
The formula for impulse (I) is given by:
I = F_avg × Δt
Where F_avg is the average force, and Δt is the time interval for which the force is exerted.
The change in momentum (Δp) is given by:
Δp = m(v_f - v_i)
Here, m is the mass of the object, v_f is the final velocity, and v_i is the initial velocity. For the ball in question:
m = 555 g = 0.555 kg (since 1 g = 0.001 kg)v_i = 14.9 m/s (initial speed, positive by convention)v_f = -12.1 m/s (final speed, negative because it is in the opposite direction after rebounding)Δt = 0.044 sFirst, we calculate the change in momentum, Δp:
Δp = 0.555 kg × (-12.1 m/s - 14.9 m/s) = 0.555 kg × (-27 m/s) = -15.285 kg·m/s
Because impulse equals change in momentum, we have:
F_avg × Δt = Δp
F_avg = Δp / Δt
F_avg = -15.285 kg·m/s / 0.044 s
F_avg ≈ -347.39 N
The negative sign indicates that the force's direction is opposite to the initial direction of the ball. However, the question asks for the magnitude, so we take the absolute value: |F_avg| ≈ 347.39 N.
A force F at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is µ. The work done by the frictional force is:a. – µFdcosθb. –Fdcosθc. –µmgdd. –µmgdcosθ
Answer:
option B
Explanation:
given,
Force = F
angle of inclination, = θ
distance = d
coefficient of friction = μ
work done =frictional Force x displacement
W = f. s
friction force on the incline will be equal to horizontal of force component. so,
f = - F cos θ
Work done will be equal to
W = - F d cos θ
Hence, the correct answer is option B
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 180,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?
Answer: The dimensions are length 600metres and breadth= 300 metres
Explanation:
Area of rectangle =l × b
L b =180,000
b = 180,000/L ....... equation 1
Perimeter of rectangle =2L + b
P = 2(180,000)/L + L
P= 360,000/L +L
dp/dL= -360,000/L^2 +1
L^2= 360,000
L= Squareroot of 360,000
L=600m
b= 180,000/600
b= 300m
The required dimensions will be "600 by 300". A complete solution is below.
According to the question,
xy = 180000No fencing along river so,
[tex]f(x) = x+2y[/tex][tex]f(x) = \frac{180000}{y+2y}[/tex]By taking the derivative, we get
→ [tex]f'(x) = -\frac{180000}{y^2+2} =0[/tex]
[tex]\frac{180000}{y^2} = 2[/tex]
[tex]y^2 = \frac{180000}{2}[/tex]
[tex]y^2 = 90000[/tex]
[tex]y = \sqrt{90000}[/tex]
[tex]= 300[/tex]
and,
→ [tex]x = \frac{180000}{300}[/tex]
[tex]= 600[/tex]
Thus the above answer is right.
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When two or more different capacitors are connected in parallel across a potential source (battery), which of the following statements must be true? (There could be more than one correct choice.)
A) The potential difference across each capacitor is the same.
B) The capacitor with the largest capacitance has the most charge.
C) Each capacitor carries the same amount of charge.
D) The capacitor with the largest capacitance has the largest potential difference across it.
Answer:
Option A and B
Explanation:
In the parallel connection, the capacitors that are connected in parallel have the same potential difference across each capacitor as the battery. Therefore, statement A is correct.
Charge (Q), capacitance (C) and voltage across capacitor (V) is related with each other as follows:
Q = C ×V
As capacitance is not same for all the capacitors, therefore each capacitor carries different charge. Therefore, statement C is incorrect.
As potential difference across each capacitor is same, therefore, capacitor with the largest capacitance carry the most charge. Therefore, statement B is correct.
Therefore, statements A and B are correct.
Answer:
Option A and B is correct
Explanation:
When two or more capacitors are connected in parallel, the equivalent capacitance of the circuit is the sum of capacitance of all the capacitors.
C = C₁ + C₂+ C₃+....
The potential difference across each capacitor is equal. Thus, option A is correct.
The total charge is the sum of the charge on stored in each capacitor.
Q = Q₁ + Q₂+ Q₃+....
Q = CV
Since the voltage is same across each capacitor, the capacitor which has largest capacitance has the most charge. Thus, option B is correct.
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is _____
Answer:
The speed of the boxes are 1 m/s.
Explanation:
Given that,
Mass of box = 1 kg
Mass of another box = 2 kg
Suppose 1 kg box moves with 3 m/s speed.
We need to calculate the speed of the boxes
Using formula of conservation of momentum
[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]
Where, u = initial velocity
v = final velocity
Put the value into the formula
[tex]1\times3+2\times0=(1+2)v[/tex]
[tex]v=\dfrac{3}{3}[/tex]
[tex]v=1\ m/s[/tex]
Hence, The speed of the boxes are 1 m/s.
After the collision, the speed of the two boxes will reduce due to lost in kinetic energy associated with inelastic collision.
The given parameters;
mass of the box, m = 1 kgmass of the second box, = 2 kgApply the principle of conservation of linear momentum to determine the final velocity of the two boxes after collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
1(u₁) + 2u₂ = v(3)
u₁ + 2u₂ = 3v
Where;
u₁ and u₂ are the initial velocity of the first and second box respectively
v is the final velocity of the two boxes
Thus, we can conclude that after the collision, the speed of the two boxes will reduce due to lost in kinetic energy associated with inelastic collision.
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A 104-hp (shaft output) electric car is powered by an electric motor mounted in the engine compartment. If the motor has an average efficiency of 91 percent, determine the rate of heat supply by the motor to the engine compartment at full load.
Answer:
10.28571 hp
Explanation:
[tex]W_{shaft}[/tex] = Shaft output power = 104 hp
[tex]\eta[/tex] = Average efficiency = 91%
Electric work done rate is given by
[tex]W_{el}=\dfrac{W_{shaft}}{\eta}[/tex]
The rate of heat supply is given by
[tex]Q=W_{el}-W_{shaft}\\\Rightarrow Q=W_{shaft}(\dfrac{1}{\eta}-1)\\\Rightarrow Q=104(\dfrac{1}{0.91}-1)\\\Rightarrow Q=10.28571\ hp[/tex]
The heat supply by the motor to the engine compartment at full load is 10.28571 hp
1 2 3 4 5 6 7 8 9 10 TIME REMAINING 59:49 Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.
Answer:
increasing the distance between the objects
decreasing the mass of one of the objects
decreasing the mass of both objects
Explanation:
The complete question would be:
Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.
increasing the distance between the objects
decreasing the distance between the objects
increasing the mass of one of the objects
increasing the mass of both objects
decreasing the mass of one of the objects
decreasing the mass of both objects
---------------------------------------------------------------------------------------
Gravitational force between two objects can be computed using the formula:
[tex]Fg = G\dfrac{Mm}{r^2}[/tex]
Where:
Fg is the gravitational force (Newtons or N)
G is the gravitational constant (6.674⋅10⁻¹¹N(m/kg)²)
M is the mass of one object (Kilograms or Kg)
m is the mass of the other object (Kilograms or Kg)
r is the distance between the two objects (m)
Now notice the relationship between the mass and gravitational force. Their relationship is direct, meaning as the value of one goes up, the other goes up and the reverse holds true. If the mass of one or both of the objects decreases, the gravitational force also decreases.
Now look at the relationship between the distance between the objects and the gravitational force-- they are indirect. This means that as one increases, the other decreases. So if we increase the distance between the objects, the gravitational force between them decreases as well.
Answer:
increasing the distance between the objects
decreasing the mass of one of the objects
decreasing the mass of both objects
Explanation:
The concentration of carbon monoxide in an urban apartment is 48 μg/m3. What mass of carbon monoxide in grams is present in a room measuring 10.0 × 10.5 × 22.5 ft?
Answer:
The answer is 3213.6 grams.
Explanation:
In meters the room dimensions are 3.05-3.2-6.86.
Total volume of the room is:
[tex]3.05*3.2*6.86=66.95[/tex]
66.95 m3.
The density of carbon monoxide is 48 g/m3. Then total grams is present in a room measuring:
[tex]66.95*48=3213.6[/tex]
Which of the following cloud services uses software to emulate hardware capabilities, such as to increase or decrease bandwidth or storage to meet changing needs?
Answer:
Iaas
Explanation:
Infrastructure as a service (IaaS) is an online service that supplies high-level APIs that are used to derefer various low-level information such as physical computing resources, location, data partitioning, scaling, protection, backup etc. of the underlying network infrastructure
Cloud computing is a cloud service that uses software to emulate hardware capabilities, such as increasing or decreasing bandwidth or storage to meet changing needs.
The cloud service that uses software to emulate hardware capabilities, such as increasing or decreasing bandwidth or storage to meet changing needs, is cloud computing.
In cloud computing, virtualization technology is used to create virtual instances of the hardware resources, allowing them to be dynamically allocated or scaled up/down as needed.
For example, Amazon Web Services (AWS) offers a cloud service called (Elastic Compute Cloud) that allows users to create virtual servers with customizable specifications, such as CPU, memory, and storage.
Users can easily increase or decrease these resources based on their requirements, without the need for physical hardware changes.
Other cloud services like Microsoft Azure, G o o g l e Cloud Platform, and IBM Cloud also provide similar capabilities, allowing organizations to optimize their resource allocation and cost efficiency.
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How does the turbine make electricity
Answer:
A wind turbine turns wind energy into electricity using the aerodynamic force from the rotor blades.
Explanation:
Suppose you go outside and look at three stars. Star A is blue, Star B is white, and Star C is red. Which star is the hottest and which star is the coldest?
Answer:
star A is hottest and star c is coldest
Explanation:
from WEIN'S LAW which relates stellar color to stellar temperature, the visible light output of very hot stars is dominated by blue color while the stars that are cool emit most of their visible light as red .
Star A, the blue star, is the hottest due to its blue color, which is a consequence of its high temperature. Meanwhile, Star C, the red star, is the coldest because red light has a longer wavelength than blue or white light.
Explanation:The color of a star can give us an idea of its temperature thanks to a concept called black-body radiation. This concept tells us that the hotter a star is, the shorter the wavelength of light it will emit, which in turn defines its color. Blue stars are the hottest, as blue light has a shorter wavelength than red or white light. As for white stars, they are intermediate in temperature, since white light is a combination of many different colors (wavelengths). Finally, red stars are the coldest of the three, because red light has a longer wavelength than blue or white light.
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A 3.8 kg block is projected at 4.4 m/s up a plane that is inclined at 37o with the horizontal. How far up along the plane does the block go:___________
(a) if the plane is frictionless and
(b) if the coefficient of kinetic friction between the block and the plane is 0.47
(c) what is thermal energy generated by friction?
Answer:
Explanation:
a) Deceleration due to gravity along the inclined plane is given by:
g*Sin(∅)= 9.8*Sin(37)=5.9 m/s^2.
Since acceleration is constant, we can use equation of motion along the plane. [tex]V^{2} = U^{2} +2as[/tex]
where V: final velocity, U:initial Velocity, a:Acceleration and s:distance
0=[tex]4.4^{2} + 2*-5.9*s[/tex] (-ve because it is against gravity)
It will go 1.64 meters up the plane
b) Force due to friction = Coeff. of friction * Normal Force
Normal Force = mg*Cos(∅) = 9.8*3.8*Cos(37) = 29.75;
Coeff of friction = 0.47;
Force due to friction = 0.47*29.75 = 13.98 N
Force due to gravity : 3.8*5.9 = 22.42;
Total force after adding friction = 22.42 + 13.98 = 36.4 N
New Acceleration = 36.4/3.8 = 9.58;
Applying the equation of motion used above with acceleration = 9.58;
It will go 1.01 meters up the plane.
c) Thermal energy due to friction = Max potential energy without friction - max potential energy after friction.
Potential energy w.o friction(m*g*h) = 3.9 * 9.8 * 1.64*Sin(37) = 37.72 Joules
Potential energy with friction(m*g*h)= 3.9 * 9.8 * 1.01*Sin(37) = 23.23 Joules
Energy due to friction = 37.72 - 23.23 = 14.49 Joules
They are adding steel components to the roof supports of the structure. We know that steel elongates when heated. At what temperature will steel elongate nine inches per 100 feet of length?
a) 500 degrees.
b) 1000 degrees.
c) 1500 degrees.
d) 2000 degrees.
Answer:
The change in temperature of 576.9°C will produce an elongation of 9 inches per feet in steel.
Explanation:
The formula for linear expansion of a material is:
ΔL = αLΔT
where, ΔL = change in length
L = Original length
ΔT = Change in temperature
α = coefficient of linear expansion
For steel, α = 13 x 10^-6 /°C
L = 100 ft
ΔL = (9 in)(1 ft/12 in) = 0.75 ft
Therefore,
0.75 ft = (13 x 10^-6 /°C)(100 ft)ΔT
ΔT = 576.9°C
A major-league pitcher can throw a baseball in excess of 44.7 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 16.1 m away from the point of release?
Explanation:
Consider the horizontal motion of ball
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 44.7 m/s
Acceleration, a = 0 m/s²
Displacement, s = 16.1 m
Substituting
s = ut + 0.5 at²
16.1 = 44.7 x t + 0.5 x 0 x t²
t = 0.36 s
Time taken to travel 16.1 m is 0.36 seconds
Now we need to find how much ball travel vertically during this 0.36 seconds.
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 0.36 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.36 + 0.5 x 9.81 x 0.36²
s = 0.64 m
The baseball drops by 0.64 meter.
After inspecting all of 55,000 kg of meat stored at the Wurst Sausage Company, it was found that 45,000 kg of the meat was spoiled. This data is described as a:A. Statistic B. Parameter
Answer:
B. Parameter.
Explanation:
After inspecting all of 55,000 kg of meat stored at the Wurst Sausage Company, it was found that 45,000 kg of the meat was spoiled.
The data above is described as a parameter because the data describes the entire population (the meat). the difference between statistic and parameters are statistic is used to describe a sample whereas parameter is used to describe the population.
The data indicating that 45,000 kg out of 55,000 kg of meat was spoiled at the Wurst Sausage Company is considered a parameter. This is because it describes the entirety of the meat stored.
Explanation:In this situation, the information about the 55,000 kg of meat, of which 45,000 kg was spoiled, is described as a parameter. A parameter is a number that describes a population, in this case, the entirety of meat stored at the Wurst Sausage Company. Whereas a statistic is a numeral that describes a sample, in this case, if they only inspected a portion of the meat storage, then the result would be a statistic.
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Two tiny beads are 25 cm apart with no other charged objects or fields present. Bead A has a net charge of magnitude 10 nC and bead B has a net charge of magnitude 1 nC. Which one of the following statements is true about the magnitudes of the electric forces on these beads?A. The force on A is 100 times the force on B.B. The force on B is 10 times the force on A.C. The force on B is 100 times the force on A.D. The force on A is 10 times the force on B.E. The force on A is exactly equal to the force on B.
Answer:
E. The force on A is exactly equal to the force on B.
Explanation:
The force between two charges is given by
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
where
[tex]q_1[/tex] = Charge on particle 1
[tex]q_2[/tex] = Charge on particle 2
r = Distance between the charges
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
[tex]F=\dfrac{8.99\times 10^9\times 10\times 10^{-9}\times 1\times 10^{-9}}{(25\times 10^{-2})^2}\\\Rightarrow F=0.0000014384\ N[/tex]
This force will be exerted on both the charges equally.
So, The force on A is exactly equal to the force on B
Two points are located on a rigid wheel that is rotating with an increasing angular velocity about a fixed axis. The axis is perpendicular to the wheel at its center. Point 1 is located on the rim, and point 2 is halfway between the rim and the axis. At any given instant, which point (if either) has the greater_______.a) angular velocity.b) angular acceleration.c) tangential speed.d) tangential acceleration.e) centripetal acceleration.
Answer:
A. The angular velocity of Point 1 and Point 2 are the same
B. The angular acceleration is equal.
C. Tangential velocity of Point 1 is greater than Point 2
D. The tangential acceleration of point1 is greater than point 2
E. The centripetal acceleration of Point 2 is greater than point 1
Explanation:
- Since angular velocity is angular displacement of Point divided by time and since they are in the same line perpendicular to the axis, that means angular velocity is equal.
The tangential speed is the measure of how quickly an object is traveling along a circular path and two object with the same angular velocity but different radius would have different tangential speed.
-The tangential acceleration is the measure of how quickly the tangential velocity changes and is angular acceleration multiplied by the radius from the centre of circle so since point 1 has a bigger radius, it would have a greater tangential acceleration.
The centripetal acceleration is the effect on the body traveling in a circular path and is directed towards the center of the circle and it is inversely proportional to the radius so the bigger the radius the smaller the centripetal acceleration.
Point 1 on the rim of the wheel has the greater tangential speed, while both points have the same angular velocity and angular acceleration.
Explanation:In this scenario, point 1, located on the rim of the wheel, has the greater tangential speed (c) since it travels a larger circle with a larger radius in the same amount of time as point 2. However, both points have the same angular velocity (a) since they are both attached to the same rigid wheel and rotate at the same rate. Additionally, both points experience the same angular acceleration (b) since they are undergoing circular motion with the same angular speed. The tangential acceleration (d) and centripetal acceleration (e) are not required to answer the question as the problem does not provide information regarding the tangential or centripetal forces acting on the points.
A Blu-ray disc is approximately 11 centimeters in diameter. The drive motor of a Blu-ray player is able to rotate up to 10,000 revolutions per minute. (a) Find the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates
Answer:
the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates is 57.6 m/s
Explanation:
Given information:
diameter of the disc, d = 11 cm, r = 5.5 cm = 0.055 m
angular speed ω = 10000 rev/min = (10000 rev/min)(2π rad/rev)(1/60 min/s)
= 1000π/3 rad/s
to calculate the maximum angular speed we can use the following formula
ω = v/r
v = ωr
= (1000π/3)(0.055)
= 57.6 m/s
A man cleaning his apartment pushes a vac-uum cleaner with a force of magnitude 42.8 N.The force makes an angle of 16.5◦with the hor-izontal floor. The vacuum cleaner is pushed2.62 m to the right along the floor.Calculate the work done by the 42.8 N force.Answer in units of J.
Answer:
107.52 J
Explanation:
force (f) = 42.8 N
angle of force (θ) = 16.5 °
distance (d) = 2.62 m
find the work done.
work done = force x distance
resolving the force we have = 42.8 x cos (16.5) = 41.04 N
work done = 41.04 x 2.62 = 107.52 J
A whistle of frequency 592 Hz moves in a circle of radius 64.7 cm at an angular speed of 13.8 rad/s. What are (a) the lowest and (b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle?
Answer:
(a) lowest frequency=577 Hz
(b) highest frequency=608 Hz
Explanation:
Given data
f(whistle frequency)=592 Hz
ω(angular speed)=13.8 rad/s
r(radius)=64.7 cm=0.647 m
To find
(a) Lowest frequency
(b) highest frequency
Solution
From Doppler effect
f=f×{(v±vd)/(v±vs)}
Where
v is speed of sound
Vd is speed detector relative to the medium(vd=0)
Vs is the speed of the source
Since
v=rω
For (a) lowest frequency
[tex]f^{i}=f(\frac{v}{v+rw} )\\f^{i}=(592Hz)(\frac{343m/s}{343m/s+(0.647m)(13.8rad/s)} )\\f^{i}=577Hz[/tex]
For (b) highest frequency
[tex]f^{i}=f(\frac{v}{v-rw} )\\f^{i}=(592Hz)(\frac{343m/s}{343m/s-(0.647m)(13.8rad/s)} )\\f^{i}=608Hz[/tex]
The engine of a 1540-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to a speed of 100 km/h at full power on a level road. Is your answer realistic?
Answer:
7.92 s
Explanation:
Power: This can be defined as the rate at which energy is used. The S.I unit of
Watt (W).
Mathematically power can be represented as,
P = E/t
Pt = E............................. Equation 1
Where P = power of the engine, t = time, E = Energy.
But,
E = 1/2m(Δv)²................. Equation 2
Where m = mass of the automobile, Δv = change in velocity of the car = final velocity - initial velocity.
Substitute the value of E in equation 2 into equation 1
Pt = 1/2m(Δv)²................................. Equation 3
making t the subject of the equation,
t = 1/2m(Δv)²/P ........................ Equation 4
Given: m = 1540 kg, P = 75 kW = 75000 W, Δv = 100-0 = 100 km/h (initial velocity of the car = 0 km/h)
100(1000/3600) m/s = 27.78 m/s
Substitute into equation 4
t = 1/2(1540)(27.78)²/75000
t = 594230.87/75000
t = 7.92 second.
Thus The time required to accelerate the car = 7.92 seconds.
It is not realistic period the time period is too short.
The car takes approximately 7.8 seconds to accelerate from rest to a speed of 100 km/h, given the power rating and mass that was provided in the question.
Explanation:The process to solve this physics problem involves steps in calculating the work done on the car and then the time required for this work at the given power. To begin with, we can translate the speed of 100 km/h to m/s, which would be 27.8 m/s. Afterward, using the kinetic energy formula (K.E = 1/2 * m * v²), the work done on the car to achieve this speed can be calculated, which is 585840 Joules.
The next step is to compute the time. We know that Power = Work Done / Time, where power has been provided (75 kW = 75000 Watts) and we just computed the work done, so we can rearrange this equation to solve for time. This gives a resulting time of approximately 7.8 seconds. This may seem a short amount of time, but it's plausible for a high-powered sports car to reach 100 km/h in around this time frame.
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Johannes Kepler used decades of Tycho Brahe's observational data to formulate an accurate description of planetary motion. Kepler spent almost 30 years of his life trying to develop a simple description of planetary motion based on a heliocentric model that fit Tycho's data. What conclusion did Kepler eventually come to that revolutionized the heliocentric model of the solar system?
Explanation:
Johannes Kepler, working with the data carefully collected by Tycho Brahe without the help of a telescope, especially those related to the retrograde motion of Mars, realized that the motion of the planets could not be explained by his model of perfect polyhedra. Coming to the conclusion that all the planets move in elliptical orbits, with the Sun in one of the foci.
A client has developed dystrophic calcification as a result of macroscopic deposition of calcium salts. The tissue that would be most affected would be?
Answer:
Tissues that are damaged or injured.
Explanation:
Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.
Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.
A person stands in a stationary canoe and throws a 5.16 kg stone with a velocity of 8.05 m/s at an angle of 31.0° above thehorizontal. The person and canoe have a combined mass of 105 kg.Ignoring air resistance and effects of the water, find thehorizontal recoil velocity (magnitude and direction) of the canoe.
magnitude m/s
direction ---Select--- opposite the horizontal component of the velocityof the stoneat right angles to the horizontal component ofthe velocity of the stonealong the horizontal component of the velocity ofthe stone
The recoil velocity of the canoe is calculated using conservation of momentum. The horizontal component of the stone's velocity is first determined, and the stone's momentum is then used to find the canoe's recoil. The direction of recoil is opposite to the stone's horizontal velocity component.
Explanation:The recoil velocity of a stationary object after projecting an object can be calculated using the principle of conservation of momentum.
Since the person and the canoe are initially motionless, the total momentum before the stone is thrown must be equal to the total momentum afterwards.
Step-by-Step Explanation
Calculate the horizontal component of the stone's velocity using the given angle and the initial velocity (vcos(31.0°)).Find the stone's momentum in the horizontal direction by multiplying its mass by the horizontal component of its velocity.Apply conservation of momentum: The momentum imparted to the stone will be equal in magnitude and opposite in direction to the momentum of the person and canoe.Calculate the recoil velocity of the canoe by dividing the stone's horizontal momentum by the mass of the person and canoe combined.The direction of the canoe's recoil will be opposite to the stone's horizontal component of velocity since momentum is conserved and no external horizontal forces are acting on the system.
An AM radio station broadcasts at a frequency of 600 kHz. If these waves have a speed of 3 x 108 m/s, then what is their wavelength in meters?
Answer: 500metres
Explanation:
Wavelength is defined as the distance between two successive crest and troughs. The relationship between the wavelength (¶), frequency (f) and velocity of the wave(v) is given by v= f¶
Therefore wavelength ¶ = v/f
Given the velocity = 3×10^8m/s
frequency = 600kHz = 600,000Hz
Wavelength =3×10^8/600,000
Wavelength = 500meters
Their wavelength in meters is 500meters
The windshield of a car has a total length of arm and blade of 9 inches, and rotates back and forth through an angle of 93degrees. What is the area of the portion of the windshield cleaned by the 7-in wiper blade?
Answer:Area cleaned by the 7-in wiper blade=22
72sq in
Explanation: In a windshield the wiper moves in a semi circle. The angle in a semi circle is 180°
Let A= Total area
A= 2×9×3.1461×(93/180)
A=29.22squareinch
Let a be area cleaned by the 7-inch wiper
a=29.22 - 2(9-7)×3.1461×(93/180)
a=29.22 - 6.502
a= 22.72sqin
Answer:
The area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²
Explanation:
Given
Length of blade = 9 inches
Angle of rotation = 93°
We're to calculate the area of the portion of the windshield cleaned by the 7-in wiper blade?
We'll solve this by using area of a sector.
Area of a sector = ½r²θ
where θ is in radians.
So, angle of rotation (93°) must first be converted to radians
Converting 93º to radians, we get 31π/60
The area of the region swept out by the wiper blade = (area of the sector where r = 9 and
θ = 31π/60) - (area of the sector where r = (9-7) and θ = 31π/60).
We're making use of 9-7 because that region is outside the boundary of the 7in blade
So Area = ½*9²*31π/60 - ½*2²*31π/60
Area = ½*31π/60(9²-7²)
Area = 31π/120 * (81 - 49)
Area = 31π/120 * 32
Area = 992π/120
Area = 62.49151386765697 in²
Area ≈ 62.49 in²
Hence, the area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²
Suppose that volumes of four stars in the Milky Way are 2.7 x 1018 km3, 6.9 x 1012 km3, 2.2 x 1012 km3, and 4.9 x 1021 km3. What is the order of the stars from least to greatest volume?
Answer:
The order is Star 3 < Star 2 < Star 1 < Star 4
Explanation:
Lets name the stars with their written order.
Star 1: 2.7 x 10^18 km3
Star 2: 6.9 x 10^12 km3
Star 3: 2.2 x 10^12 km3
Star 4: 4.9 x 10^21 km3
Star with lowest power of 10 has the least volume. Therefore, star 2 and star 3 would be the stars with least volume. Star 2's coefficient is better than Star 3. Thus, the order will be Star 3 < Star 2 < Star 1 < Star 4.
The belt of one ramp moves at a constant speed such that a person who stands still on it leaves the ramp 64 s after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of 0.37 m/s2, he cov- ers the same distance as the ramp does, but in one-fourth the time. What is the speed at which the belt of the ramp is moving?
Answer:
0.74 m/s
Explanation:
given,
time taken by the ramp to cover cliff , t = 64 s
acceleration = 0.37 m/s²
distance traveled by the belt
[tex]x = v_{belt}t_{belt}[/tex]
Clifford is moving with constant acceleration
[tex]x = v_ot + \dfrac{1}{2}at^2[/tex]...(1)
initial velocity is equal to zero
[tex]x =\dfrac{1}{2}at^2[/tex]........(2)
equating equation (1) and (2)
[tex]v_{belt}t_{belt}=\dfrac{1}{2}at^2[/tex]
[tex]v_{belt}t_{belt}=\dfrac{1}{2}a(\dfrac{1}{4}\times t_{belt})^2[/tex]
[tex]v_{belt}=\dfrac{1}{32}a t_{belt}[/tex]
[tex]v_{belt}=\dfrac{1}{32}\times 0.37 \times 64[/tex]
[tex]v_{belt}= 0.74\ m/s[/tex]
Speed of the belt is equal to 0.74 m/s
Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?
Answer:
0.83999 m
0.20999 m
Explanation:
g = Acceleration due to gravity = 9.81 m/s² = a
s = 189 cm
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s[/tex]
When the time intervals are equal, if four drops are falling then we have 3 time intervals.
So, the time interval is
[tex]t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s[/tex]
For second drop time is given by
[tex]t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s[/tex]
Distance from second drop
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m[/tex]
Distance from second drop is 0.83999 m
Distance from third drop
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m[/tex]
Distance from third drop is 0.20999 m
A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m below the surface. Find the net force exerted on the diver while in the water.
Answer:
(a) 13.795 m/s.
(b) -3140.28 N.
Explanation:
(a) Using newton's equation of motion,
v² = u² + 2gs.......................... Equation 1
Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.
Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².
Substitute into equation 1
v² = 0² + 2×9.81×9.7
v² = 190.314
v = √(190.314)
v = 13.795 m/s.
Hence the velocity of the driver when he hits the water = 13.795 m/s.
(b)
F = ma.................... Equation 2
Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.
Also using
v² = u² + 2as ............ Equation 3
Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s
Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s
Substitute into equation 3
0² = 13.795²+2(2a)
0 = 190.30203 + 4a
-4a = 190.30203
a = 190.30203/-4
a = -47.58 m/s²
Also given: m = 66 kg,
Substitute into equation 3
F = (-47.58)(66)
F = -3140.28
Note: The Force is negative because it act against the motion of the diver.
Hence the net force exerted on the diver while in the water = -3140.28 N.