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padilas [110]
3 years ago
6

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?
Physics
1 answer:
dsp733 years ago
4 0

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

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Ksivusya [100]
<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

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                       M = Mass of body 1

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                       r = Distance between them

Here we have

                 M = Mass of Sun = 1.99×10³⁰ kg

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                 F = 3.14×10¹³ N

Substituting

                   F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m

The asteroid is 4.11 x 10¹¹ m far from Sun

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3 years ago
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son4ous [18]
The most common liquid on planet earth is water
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Answer:

Explanation:

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Put the value in the equation (I)

v=\dfrac{3\times10^{8}}{1.63}

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