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Naddika [18.5K]
4 years ago
15

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma

ss m, moves with velocity (-29 m/s) and a second piece, also of mass m, moves with velocity (-29 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the +x axis) of the velocity of the third piece?
Physics
1 answer:
Brilliant_brown [7]4 years ago
3 0

Answer with Explanation:

We are given that

m_1=m_2=m

v_1=-29 im/s

v_2=-29 jm/s

m_3=3m

a.Initial velocity of vessel,u=0

According to law of conservation of momentum

Mu=m_1v_1+m_2v_2+m_3v_3

Where M=Mass of  vessel

0=m(-29i)+m(-29j)+3mv_3

29im+29jm=3mv_3

v_3=29i+29jm/s

Magnitude,\mid v_3\mid=\sqrt{(29)^2+(29)^2}=29\sqrt 2m/s

b.\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{29}{29})=45^{\circ}

Hence, direction of the velocity of third piece=45 degree

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Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

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After comparing this expression with the first one, the following equivalence is found:

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g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

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The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

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