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spayn [35]
2 years ago
12

The passengers are facing forward in the direction that the train is moving. What will happen to the passengers sitting on a tra

in if the train comes to a sudden stop? fall Question 13 options: They will feel a force pushing them downward. They will feel the force pushing them upward. Their bodies will move backwards. Their bodies will move forward.
Physics
1 answer:
marysya [2.9K]2 years ago
6 0

Passengers which are facing forward in the direction that train is moving, their bodies will move forward in application of sudden stop of train.

<h3>What is force of inertia?</h3>

Force of inertia is the force which acts in the opposite direction of the force of acceleration acting on the body.

Given infroamtion-

The passengers are facing forward in the direction that the train is moving.

The train comes to a sudden stop.

Lets see what happens step wise-

  • Here, the train in moving in the forward direction and the passengers are also facing forward in the direction that the train is moving.

  • Now the train comes to a sudden stop. By this sudden stop the train stops suddenly but all the object including the passengers is still travelling forward due to the inertia force.

  • Thus all the passenger will tend to move in the direction as they are still travelling.

Hence, passengers which are facing forward in the direction that train is moving, their bodies will move forward in application of sudden stop of train.

Learn more about the force of inertia here;

brainly.com/question/10454047

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AFTER the bend, since the light rays then travel in a different direction, we may also say that the 'velocity' has changed.

8 0
3 years ago
Please help <br> Convert the following distance v time graph velocity graph
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5 0
2 years ago
The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

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k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0
3 years ago
What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters
Volgvan
A_central = v^2/r = (19)^2/10 = 36.1 m/s^2
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dmitriy555 [2]
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How demands became the obstacles ?

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