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bezimeni [28]
4 years ago
9

Which has more momentum, a speeding baseball or an ocean liner at rest in a harbor?

Physics
1 answer:
uranmaximum [27]4 years ago
4 0
Momentum is (mass) times (speed), so nothing that is at rest has any momentum. If the battleship is at rest, then a mosquito in flight, a leaf falling from a tree, and your speedy baseball each have more momentum than the ship has.
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A train is pulling four train cars and each car has a mass of 40,000 kg. The train is accelerating at 1.1 m/s^2. What is the for
IgorLugansk [536]

Answer:

176,000 N

Explanation:

Newton's second law:

∑F = ma

F = (4 × 40,000 kg) (1.1 m/s²)

F = 176,000 N

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4 years ago
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Why does a diverging lens never produce a real image?
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A diverging lens never produces a real image because the actual light rays never converge. They always diverge. ... A diverging virtual image is always SMALLER than the object. A converging virtual image is always LARGER than the object
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If Ike notices that there is a new moon tonight, when should he expect there to be a new moon again?
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Charlie is investigating friction. She will use the same amount of force to push two wooden balls across two level surfaces. The
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D the distance for trial 3 will be greater than trial 4
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A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia
Ierofanga [76]

Answer:

v_{0}=319.2 m/s    

Explanation:

We need to use the momentum and energy conservation.

p_{0}}=p_{f}

mv_{0}=(m+M)V_{1}

Where:

  • m is the mass of bullet (m=0.01 kg)
  • M is the mass of the wooden (M=0.75 kg)
  • v(0) initial velocity of bullet
  • V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.            

(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh

Here:

  • V(1) is the velocity of the combined at the initial position
  • h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

cos(\theta) =(L-h)/L=(1.72-0.8)/1.72

\theta =57.66

Now, we can solve the equation to find V(2)

V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}

V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}

V_{2}=1.40 m/s        

Now we can find V(1) using the conservation of energy equation

(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh

V_{1}=\sqrt{V_{2}^{2}+2gh}

V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}          

V_{1}=4.20 m/s        

Finally, using the momentum equation we find v(0)

v_{0}=\frac{(m+M)V_{1}}{m}                

v_{0}=\frac{(0.01+0.75)*4.20}{0.01}

v_{0}=319.2 m/s        

I hope it helps you!

 

7 0
4 years ago
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