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iragen [17]
3 years ago
12

a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o

f 4.20 m. what is the magnitude of the force applied by the brakes?(unit=N) PLEASE HELP
Physics
2 answers:
Ratling [72]3 years ago
8 0

Answer:

4,955.9 N

Explanation:

1/2 (m*v^2) = 1/2 (1210 x 6.33^2) = 24,241.7 J.

1/2 (1210 x 2.38^2) = 3,427 J.

(24,241.7 - 3,427) = 20,814.7J. of work done by the brakes.

(20,814.7/ 4.2m) = brake force of 4,955.9 N.

Stels [109]3 years ago
4 0

Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

F = ma

F = (1210 kg) (-4.10 m/s²)

F = -4960 N

The magnitude of the force is 4960 N.

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b)

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A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

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We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

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Learn more: brainly.com/question/16274121

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