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Bingel [31]
3 years ago
11

A 0.30-kg object is traveling to the right (in the positive direction) with a speed of 3.0 m/s. After a 0.20 s collision, the ob

ject is traveling to the left at 4.0 m/s. What is the magnitude of the impulse (in N-s) acting on the object
Physics
1 answer:
andre [41]3 years ago
5 0

Answer:

I = -2.1 kg.m/s

Explanation:

Given,

mass of the object,m = 0.30 Kg

initial speed, v_i = 3 m/s

time of collision = 0.20 s

final speed, v_f = -4 m/s

Impulse = change in momentum

I = m (v_f-v_i)

I = 0.30\times (-4-3)

I = -2.1 kg.m/s

Hence, impulse of the object is equal to I = -2.1 kg.m/s

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Parallel to the handle, what is the force of friction acting on the suitcase?
mrs_skeptik [129]
 <span>Lets call F the friction force which will act horizontally backwards. 

As you are travelling at a constant velosity horizontally there is no overall resultant force in this direction. 

ie. the force you pull with will be equal to the friction force resisting you. (you will initially have to have pulled with a greater force than the friction to get the suitcase moving) 

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A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
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a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

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h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

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b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

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