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IceJOKER [234]
3 years ago
5

A hunter points a rifle directly at a coconut that he intends to shoot off a tree. It so happens that the coconut falls from the

tree at the exact instant the hunter pulls the trigger. What happens to the bullet?
Show your motion diagram and complete solution.

Show the solution to prove your answer.

It passes above the coconut

It passes beneath coconut

It hits the coconut
Physics
1 answer:
yan [13]3 years ago
4 0

Answer:

It hits the coconut

Explanation:

The coconut fell immediately after the trigger was pulled as a result of the impact made by the bullet on the coconut. The Third Law of  motion states that : For every action, there is an equal and opposite reaction. The action was performed when the bullet struck the coconut, while the coconut reacted by falling of the tree.

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The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.
Alexus [3.1K]

Answer:

(a) θ= 43.89°

(b) v_{1} = 1.88\sqrt{3} \frac{m}{s}

    v_{2} = 6.79 \frac{m}{s}

Explanation:

Ball 1:

u_{1} = 7.52\frac{m}{s}

Ball 2:

u_{2} = 0\frac{m}{s}

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

\frac{1}{2} m_{1}u_{1}^{2}   = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}

and

m_{1}u_{1} =   m_{1}v_{1} + m_{2}v_{2}

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

u_{1}^{2}   = v_{1}^{2} + v_{2}^{2}

and

u_{1} =   v_{1} + v_{2}

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

56.55 = v_{1} ^{2} + v_{2} ^{2}

and

u_{1} = v_{1}cos(30) +  v_{2}cos(\theta)

Solving we get:

(\frac{2u_{1}-\sqrt{3}v_{1}  }{2cos(\theta)})^{2}  = 56.55-v_{1} ^{2}

From conservation in y-direction, we get:

0 = v_{1}sin(30) -  v_{2}sin(\theta)

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

8 0
3 years ago
How much force is needed to stop a 50 kg gymnast if he decelerates at 25 m/s^2
vlada-n [284]
The force can be calculated by multiplying the mass of the gymnast with her acceleration.

Force = 50 kg × 25 m/s2
Force = 1250 N

A force of at least 1250 N can stop the 50-kg gymnast.

I hope I was able to answer your question. Have a good day.
3 0
3 years ago
When entering an expressway, in the acceleration lane you should: A search for a gap in traffic and adjust your speed to the spe
sattari [20]

Answer:A

Explanation: search for a gap in traffic and adjust your speed to the speed of the traffic.

3 0
3 years ago
The weight of an object is the product of its mass, m, and the acceleration of gravity, g (where g=9.8 m/s2). If an object’s mas
lina2011 [118]

You just told us that  Weight = (mass) x (gravity),
and that mass=10kg and  gravity = 9.8 m/s².

All we need to do is write those numbers in place of the letters.

                     Weight = (10 kg) x (9.8 m/s²)  =  <em>98 newtons</em>.

6 0
4 years ago
Given s=1/2(v+u)t find t ​
Andrej [43]

Explanation:

s=(v+u)t/2

2s/(v+u)=t

.................

6 0
3 years ago
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