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astra-53 [7]
3 years ago
11

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.

Physics
2 answers:
Aloiza [94]3 years ago
8 0

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:

F=ma

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.

MrRissso [65]3 years ago
3 0

Answer:

I’s B

Explanation:

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.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum accelerat
AlexFokin [52]

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg  

assume the greatest scale reading = 78.09 kg

<u>Determine the maximum acceleration for these elevators</u>

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u> </u>

<u> </u>At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

3 0
3 years ago
A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina
elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec        

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car a=1.6m/sec^2 in negative direction so acceleration will be a=-1.6m/sec^2

From first equation of motion we know that

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So v=0+(-1.6)\times 5.8=-9.28m/sec

So final velocity will be -9.28 m/sec

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A 47.2 kg girl is standing on a 177 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
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Answer:

v_g,i = 1.208 m/s

Explanation:

We are given;

Mass of girl; m_g = 47.2 kg

Mass of plank; m_p = 177 kg

Let the velocity of girl to ice be v_g,i

Let the velocity of plank to ice be v_p,i

Since the velocity of the girl is 1.53 m/s relative to the plank, then;

v_g,i + v_p,i = 1.53

From conservation of momentum;

m_g × v_g,i = m_p × v_p,i

Thus;

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Dividing both sides by 47.2 gives;

v_g,i = 3.75(v_p,i)

v_pi = (v_g,i)/3.75

Thus, from v_g,i + v_p,i = 1.53, we have;

v_g,i + ((v_g,i)/3.75) = 1.53

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v_g,i = 1.53/1.267

v_g,i = 1.208 m/s

5 0
3 years ago
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