Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr
iving at steady speed up a 10∘ incline. C. A car speeding up after leaving a stop sign. D. A car driving at steady speed around a curve.
2 answers:
Answers:
A. A car driving at steady speed on a straight and level road.
B. A car driving at steady speed up a 10∘ incline.
Explanation:
An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:
Since F=0, a=0 as well.
Let's now analyze each case.
A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.
D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.
So the only two choices which are correct are A and B.
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Answer:
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Explanation:
There are two (2) conditions, when the desk is about to move and when the desk is moving. In the attachements you can see the two free body diagram for each condition.
In the first condition, there is no movement and the force is 12 N, in the image we can see the total forces are equal to 0 and by the definition of the friction force we can get the static friction coefficient.
In the second condition there is movement in the direction of the force which is equal to 8 N, again by the definition of the friction force we can get the kinetic friction coefficient. Since the desk is moving with constant velocity there is not acceleration.
Answer:
0.075 m
Explanation:
The picture of the problem is missing: find it in attachment.
At first, block A is released at a distance of
h = 0.75 m
above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:
where
is the acceleration due to gravity
is the mass of the block
is the speed of the block A just before touching block B
Solving for the speed,
Then, block A collides with block B. The coefficient of restitution in the collision is given by:
where:
e = 0.7 is the coefficient of restitution in this case
is the final velocity of block B
is the final velocity of block A
is the initial velocity of block B
Solving,
Re-arranging it,
(1)
Also, the total momentum must be conserved, so we can write:
where
And substituting (1) and all the other values,
This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to
where
k = 600 N/m is the spring constant
x is the compression of the spring
And solving for x,
Answer: Tension = 47.8N, Δx = 11.5× m.
Tension = 95.6N, Δx = 15.4× m
Explanation: A speed of wave on a string under a tension force can be calculated as:
is tension force (N)
μ is linear density (kg/m)
Determining velocity:
0.0935 m/s
The displacement a pulse traveled in 1.23ms:
Δx = 11.5×
With tension of 47.8N, a pulse will travel Δx = 11.5× m.
Doubling Tension:
|v| = 0.1252 m/s
Displacement for same time:
15.4×
With doubled tension, it travels 15.4×
m