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Anastasy [175]
2 years ago
15

A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4. 5 m/s. What is the car’s change in momentum d

ue to this collision?.
Physics
1 answer:
kodGreya [7K]2 years ago
4 0

Answer:

Explanation:

Change in car's momentum = 700 * [4.5 - {-29)] = 23,450 kgm/s

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An equinox occurs when
schepotkina [342]
A! Good luck on your test!
6 0
3 years ago
Read 2 more answers
An object experiences an acceleration of -6.8m/s^2. As a result, it accelerates from 54m/s to a complete stop. how much distance
Kipish [7]

Answer:

210 m

Explanation:

Given:

a = -6.8 m/s²

v₀ = 54 m/s

v = 0 m/s

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (54 m/s)² + 2 (-6.8 m/s²) Δx

Δx ≈ 210 m

6 0
3 years ago
A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of i
kirza4 [7]

Answer:

2.5 x 10⁷ J

Explanation:

F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87  m

Work done by the engine is given as

W = F d =  (2.3 x 10⁵) (87) = 200.1 x 10⁵ J

W' = Net work done

W'' = work done by catapult

KE₀ = initial kinetic energy = 0 J

KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

We know that

W' = W + W''

4.5 x 10⁷ = 2.001 x 10⁷ + W''

W'' = 2.5 x 10⁷ J

8 0
3 years ago
What theory is used to explain the behavior of particles in gases?
KonstantinChe [14]

kinetic molecular theory


4 0
3 years ago
Read 2 more answers
Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to b
weeeeeb [17]

Answer:

4.408 m/s, 4.102 m/s, 4.026 m/s

Explanation:

The question is incomplete. The text of the original question states:

A race car moves such that its position fits the relationship

:

x=(4.0 m/s)t + (0.85 m/s^3) t^3

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.

We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s

5 0
3 years ago
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