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Charra [1.4K]
3 years ago
10

A motorcycle has a mass of 2.50×10^2. A constant force is exerted on it for 60.0s. The motorcycles initial velocity is 6.00 m/s

and its velocity 28.0 m/s. What is it’s change in momentum?
Physics
1 answer:
Margarita [4]3 years ago
8 0

The change in momentum is 5500 kg m/s

Explanation:

The change in momentum of an object is given by

\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have:

m=2.50\cdot 10^2 kg (mass of the motorcycle)

v=28.0 m/s (final velocity)

u=6.0 m/s (initial velocity)

Therefore, the change in momentum is

\Delta p=(2.50\cdot 10^2)(28.0-6.0)=5500 kg m/s

Learn more about change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

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Answer:

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b) See the attached figure. Slope of the line = 32.

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Explanation:

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v = ΔX / Δt

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Considering the origin of the reference system as the position where the observer is:

ΔX = 210 m - 50 m = 160 m

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b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:

slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.

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A 1.40-kg ball tied to a string fixed to the ceiling is pulled to one side by a force F→ . where L = 1.40 kg. What is the tensio
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