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guapka [62]
3 years ago
11

A 3.1-kilogram gun initially at rest is free to

Physics
2 answers:
34kurt3 years ago
8 0
The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.

The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:

7.5=3.1x
x=7.5/3.1
x=2.42

So the speed of the gun is 2.4m/s.
Svetradugi [14.3K]3 years ago
8 0
     Let us consider the gun with the index 1 and bullet with the index 2. Using the Momentum consevation Equation, we have:

\Delta Q= 0 \\ m_{1}*v_{1}-m_{2}*v_{2}=0 \\ m_{1}*v_{1}=m_{2}*v_{2}
 
     Entering the unknown, comes:

m_{1}*v_{1}=m_{2}*v_{2} \\ 3.1*v_{1}=0.015*500 \\ v_{1}= \frac{7.5}{3.1} \\ \boxed {v_{1}=2.4m/s}

Number 2

Obs: approximate results.

If you notice any mistake in my english, please let me know, because i am not native.
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Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
  • \alpha = angle of the final position vector from the observation point = 123^\circ
  • \vec{AB} = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

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3 years ago
If a machine has an efficiency of 94% and you apply 574J of work, how much work do you get out of the machine
Mariana [72]
<h3>Answer:</h3>

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<h3>Explanation:</h3>
  • Efficiency of a machine is the ratio of work output to work input expressed as a percentage.
  • Efficiency = (work output/work input) × 100%
  • Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.

In this case;

Efficiency = 94%

Work input = 574 Joules

Therefore, Assuming work output is x

94% = (x/574 J) × 100%

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