What happens in terms of energy when a moving car hits a parked car, causing the parked car to move?
Answers
ANSWER:
The moving car transfers kinetic energy to the parked car.
EXPLANATION:
Kinetic energy is the energy an object has because of its motion and can be transferred between objects and transformed into other kinds of energy.
When a moving car hits a parked car, causing the parked car to move, the type of collision is elastic collision. An elastic collision is when two bodies collide and separates after collision conserving the total kinetic energy before and after collision.
Related Questions
A motorboat travels directly east. During its trip, winds blow from directly north, directly south, and southwest. Which of the following statements is true? A. All of the winds do work on the boat. B. Only the north wind does work on the boat. C. Only the south wind does work on the boat. D. Only the southwest wind does work on the boat.
Researchers studying __________ observe behavioral changes in people with damage to specific brain areas.
A helicopter flying west begins experiencing an acceleration of 3m/s2 east. Will the magnitude of its velocity increase or decrease?
What happens to parallel light rays that strike a concave lens?
b. the light
c. the wave
d. the speed
Answers
For refraction to occur, there has to be change from one medium to another. For example rays of light would refract as the travel from air to water.
B. 0.4275 m
C. 0.6125 m
D. 0.8000 m
Answers
Answer
C. 0.6125 m
Explanation
The magnification of an image produced by curved mirrors can be given in two ways;
Magnification, M = (Height of the image) / (height if the object) and
M = (image distance,V) / (Object distance, U).
Using the first formula,
M = 0.35/0.4
= 0.875
Now from the second formula,
M = V/U
0.875 = V/0.7
V = 0.875 × 0.7
= 0.6125 m
Answer : v = 0.1625 m.
Explanation :
It is given that,
Height of an object,
Object distance,
Height of an image,
We have to find the image distance (v).
We know that the magnification of mirror is the ratio of height of image to the height of object.
i.e. ..............(1)
Also, magnification is the ratio of image distance to that of object distance.
i.e ............(2)
So, from equation (1) and (2)
So, image is located at a distance of 0.6125 m.
(b Can this plane land on a runaway that is only 0.800 km long?
shown work pls will reward alot of points
Answers
Answer:
a) t = 20 s, b) x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance
Explanation:
This is a kinematics exercise
a) in minimum time to stop,
v = vo + at
v = 0
t = -v0 / a
we calculate
t = -100 / (5.00)
t = 20 s
b) Let's find the length you need to stop
v² = vo² + 2 a x
x = -v0 ^ 2 / 2a
x = - 100² / 2 (-5.00)
x = 1000 m
As the runway is only 800 m long, the plane cannot land at this distance.
Answers
Answer:
the velocity of a moving body relative to another body is called relative velocity.
Answers
1. How fast is the blue car going 1.8 seconds after it starts?
Recall this kinematic equation:
Vf = Vi + aΔt
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest)
a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)
Δt = 1.8 s
Substitute the terms in the equation with the given values and solve for Vf:
Vf = 0 + 3.7×1.8
Vf = 6.66 m/s
2. How fast is the blue car going 10.0 seconds after it starts?
The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.
We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:
Vf = 0 + 3.7×4.4
Vf = 16.28 m/s
The constant velocity at which the car is traveling at t = 10s is 16.28 m/s
3. How far does the blue car travel before its brakes are applied to slow down?
We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.
Traveling under constant acceleration:
Recall this kinematic equation:
d = ×Δt
d is the distance traveled.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 16.28 m/s (determined from question 2).
Δt = 4.4 s
Substitute the terms in the equation with the given values and solve for d:
d = ×4.4
d = 35.8 m
Traveling at constant velocity:
Recall the relationship between velocity and distance:
d = vΔt
d is the distance traveled.
v is the velocity.
Δt is the amount of elapsed time.
Given values:
v = 16.28 m/s (the constant velocity from question 2).
Δt = 8.3 s (the time interval during which the car travels at constant velocity)
Substitute the terms in the equation with the given values:
d = 16.28×8.3
d = 135.1 m
Add up the distances traveled.
d = 35.8 + 135.1
d = 170.9 m
4. What is the acceleration of the blue car once the brakes are applied?
Recall this kinematic equation:
Vf²=Vi²+2ad
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration
d is the distance traveled.
Given values:
Vi = 16.28 m/s
Vf = 0 m/s
d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)
Substitute the terms in the equation with the given values and solve for a:
0² = 16.28²+2a×45.1
a = -2.94 m/s²
5. What is the total time the blue car is moving?
We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.
Recall again:
d = ×Δt
Given values:
d = 45.1 m
Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).
Vf = 0 m/s (the car slows to a stop).
Substitute the terms in the equation with the given values and solve for Δt:
45.1 = ×Δt
Δt = 5.54s
Add up the times to get the total travel time:
t = 4.4 + 8.3 + 5.54 =
t = 18.24s
6. What is the acceleration of the yellow car?
Recall this kinematic equation:
d = ViΔt + 0.5aΔt²
d is the distance traveled.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
d = 216 m (both cars meet at 216m)
Vi = 0 m/s (the car starts at rest)
Δt = 18.24 s (take the same amount of time to reach 216m)
Substitute the terms in the equation with the given values and solve for a:
216 = 0×18.24 + 0.5a×18.24²
a = 1.3 m/s²
Answers
The probability of (E and F) = P(e) times P(f) = 0.76 x 0.45 = 0.342 .
The probability of (e OR f) =
1 - P(not e AND not f) =
1 - P(not e) x P(not f) =
1 - (1 - 0.76) x (1 - 0.32) =
1 - ( 0.24 ) x ( 0.68 ) = 1 - 0.1632 = 0.8368.