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antiseptic1488 [7]
3 years ago
6

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

f a gully. Its momentum upon hitting the ground is
A) 1,000 kg m/s.
B) 2,000 kg m/s.
C) 3,000 kg m/s.
D) 4,000 kg m/s.
E) 9,000 kg m/s.
Physics
1 answer:
Jet001 [13]3 years ago
8 0

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

v(t)=v_0 -gt

where

v_0=0 is the initial vertical velocity (zero, since the motorcycle is not moving)

g = 9.8 m/s^2 is the acceleration due to gravity

t is the time

Since the motorcycle hits the ground after t = 3 seconds, we have

v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s

And since we know its mass, m=100 kg, we can find its momentum:

p=mv=(100 kg)(-29.4 m/s)=-2940 kg m/s \cdot -3000 kg m/s

and the negative sign simply means downward direction.

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Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

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Select one:

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Answer:

b. 1.3 s

Explanation:

Given;

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when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

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If a particle undergoes shm with amplitude 0.21 m, what is the total distance it travels in one period?
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The ransom note in the Lindbergh kidnapping case included some obvious spelling errors and other foreign and regional language.
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A sailboat runs before the wind with a constant speed of 2.8 m/s in a direction 52° north of west. How far (a) west and (b) nort
vodka [1.7K]
<h2>Displacement along west = 3612 m</h2><h2>Displacement along north = 4633.50 m</h2>

Explanation:

Let east be positive x axis and north be positive y axis

Velocity of boat = 2.8 m/s in a direction 52° north of west.

Velocity, v = -2.8 cos 52 i + 2.8 sin 52 j = -1.72 i + 2.21 j m/s

Time taken = 35 min = 35 x 60 = 2100 s

Displacement = Velocity x Time

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Displacement =  -3612 i + 4633.50 j m

Displacement along west = 3612 m

Displacement along north = 4633.50 m

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