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lyudmila [28]
3 years ago
8

During a circus act, a performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of m1 = 7.5

kg and the horizontal component of its velocity is v = 8.85 m/s when the performer, of mass m2 = 61.5 kg, catches it.
If the performer is on nearly frictionless roller skates, what is his recoil velocity?
Physics
1 answer:
vladimir1956 [14]3 years ago
3 0

Answer:

Explanation:

We shall apply law of conservation of momentum .

total Initial momentum =

7.5 x 8.85 = 66.375 kg m / s

total mass = 7.5 + 61.5 = 69 kg , common velocity of ball and performer be v

Total momentum = 69 v

69 v  = 66.375

v = 66.375 / 69

= .962 m /s .

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wlad13 [49]

Answer:

a) V = 0.82m/s

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Explanation:

By conservation of energy we know that:

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Solving for V we get:

V = 0.82 m/s

To find the maximum speed we will do the same to an intermediate point where the compression is X and the distance for the work donde by frictions is given by (Xmax - X) = (0.28m - X):

\frac{m*Vmax^{2}}{2}+\frac{K*X^{2}}{2}-\frac{K*Xmax^{2}}{2}=-Ff*(Xmax-X)

Then we have to solve for V, derive and equal zero in order to find position X. After solving the derivative we get:

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6 0
3 years ago
How far from the surface of Earth is the magnitude of Earth's gravitational field equal to 7.86 N/kg?
AfilCa [17]

Answer:

7.48 x 10⁵ m

Explanation:

g = 7.86 N/kg

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R + h = 7.12 x 10⁶ m

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