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vivado [14]
3 years ago
10

A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o

f Callisto is 1.08 × 1023 kg. The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero kg · m/s, what is the recoil speed of Callisto immediately after the collision?
Physics
1 answer:
vredina [299]3 years ago
7 0

Answer:

Recoil speed, 1.17\times 10^{-5}\ m/s                          

Explanation:

Given that,

Mass of the comet fragment, m_1=1.96\times 10^{13}\ kg

Speed of the comet fragment, v_1=6.5\times 10^4\ m/s

Mass of Callisto, m_2=1.08\times 10^{23}\ kg

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

m_1v_1=(m_2+m_2)V

V is recoil speed of Callisto immediately after the collision.

V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s

So, the recoil speed of Callisto immediately after the collision is 1.17\times 10^{-5}\ m/s

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Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

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8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

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Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

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9.28 + 8.3 x = 0

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5 0
3 years ago
If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that
alexandr402 [8]

Answer:

g'=63.74\ m/s^2

Explanation:

It is given that,

Weight of the person on Earth, W = 818 N

Weight of a person is given by the following formula as :

W=mg

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m=\dfrac{W}{g}

m=\dfrac{818\ N}{9.8\ m/s^2}

m = 83.46 kg

The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

g'=\dfrac{W'}{m}

g'=\dfrac{5320\ N}{83.46\ kg}                

g'=63.74\ m/s^2

So, the acceleration due to gravity on that planet is 63.74\ m/s^2. Hence, this is the required solution.                                                                    

6 0
3 years ago
1. Which of the following statements best describes matter?
Rashid [163]

Answer:

Matter is anything that has mass

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The word "matter" refers to anything that has mass, either organic or inorganic. Matter is made up of atoms, which consists of a nucleus (made up of protons, positively charged, and neutrons, electrically neutron) and electrons which revolve around the nucleus.

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Matter can be in three different states also:

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3 years ago
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Maurinko [17]

Answer:

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Using equation of motion

v^2=u^2+2gs

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

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u=\sqrt{2\times9.8\times11.0}

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Hence, The speed of the water is 14.68 m/s.

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Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

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Because the work done = Force x Distance

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3 0
3 years ago
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