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andriy [413]
1 year ago
14

A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 32 m at t2 = 2.0 s .

Answer the following by considering ratios, without computing the car's velocity. 1. What is the car's position at t = 1.0 s ? 2. What will be its position at t = ? answers pleaseeeee
Physics
1 answer:
Lerok [7]1 year ago
3 0

Without computing the car's velocity, the car's position at t = 1.0 s will be x = 16m

<h3>Position - Time Graph</h3>

The gradient of the position time graph is speed.

Given that a car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 32 m at t2 = 2.0 s.

The gradient m = Δx ÷ Δt

m = (32 - 0) / (2 - 0)

m = 32 / 2

m = 16 m/s

Without computing the car's velocity, the car's position at t = 1.0 s will be

32 / 2 = x / 1

x = 16 m

Therefore, without computing the car's velocity, the car's position at t = 1.0 s will be x = 16m

Learn more about Position Time Graph here: brainly.com/question/18388330

#SPJ1

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A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re
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Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

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v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

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Answer:

The answer is 4200 J.

Explanation:

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