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timurjin [86]
3 years ago
12

If you push on an object that has a mass of 400 kg and it accelerates at 4 m/s2, what would the acceleration (a) be if the mass

was reduced to 1/4 as much?
(Hint: Instead of plugging 400 kg into the equation, plug 100 kg into the equation as the mass but keep the force at 1,600 N.)

a
1/4 m/s2

b
16 m/s2

c
8 m/s2

d
4 m/s2
Physics
1 answer:
Temka [501]3 years ago
6 0

Answer:

b  16 m/s².

Explanation:

Hello.

In this case, the force exerted to the mass of 400 kg at an acceleration of 4 m/s² is:

F=400kg*4m/s^2=1600N

Now, since the mass is decreased to 100 kg (1/4 of 400 kg), and the force remain the same, the new acceleration should be:

a=\frac{F}{m}=\frac{1600kg*m/s^2}{100kg}\\  \\a=16m/s^2

Thus, the answer is b  16 m/s².

Best regards.

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The momentum of the rocket is

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From eq.(1) we get

v=\sqrt{\frac{2K}{m}}

and substituting into (2),

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Now in this problem we have:

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K' = 8K

- The mass is reduced by half:

m'=\frac{m}{2}

Substituting, we find the new momentum:

p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p

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4 0
3 years ago
A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi
Marrrta [24]

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h = 2.64 meters      

Explanation:

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Speed of the first ball, v_1=20\ m/s (upward)

Mass of the other ball, m_2=2\ kg

Speed of the other ball, v_2=-12\ m/s (downward)

We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :

V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}

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mgh=\dfrac{1}{2}mV^2

h=\dfrac{V^2}{2g}

h=\dfrac{7.2^2}{2\times 9.8}

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5 0
3 years ago
A ball is dropped from a height h and falls the last half of its distance in 4 seconds. How long does the ball fall? From what h
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Answer:

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the atlantic torpedo is a large electric fish capable of generating a voltage of 220V between its tail and its head. This drives
Elza [17]
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Korvikt [17]

Answer:

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7 0
3 years ago
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