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Artyom0805 [142]
3 years ago
7

. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be

tween child and ice, find the recoil velocity of the child
Physics
1 answer:
tamaranim1 [39]3 years ago
6 0

Answer:

Explanation:

A 40kg child throw stone of 0.5kg

At a direction of 5m/s

Recoil can be calculated using recoil of a gun formula

m_1•v_1 + m_2•v_2

m_1•v_1 = -m_2•v_2

The negative sign show that the momentum of the boy is directed oppositely to that of the stone

m_1 Is mass of boy

v_1 is the recoil velocity of the boy

m_2 is mass of stone

v_2 is the velocity of stone

Then,

m_1•v_1 = -m_2•v_2

40•v_1 = -0.5 × 5

40•v_1 = -2.5

v_1 = -2.5 / 40

v_1 = -0.0625 m/s

The recoil velocity of the boy is 0.0625 m/s

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Answer:

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b)x=2.405 m

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The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

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F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

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b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

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\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

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