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Kobotan [32]
2 years ago
7

He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle

t lodges itself inside the tree stump before the system of objects slides along the ice. What is the speed of the tree stump-bullet system immediately after the collision?
Physics
1 answer:
Ymorist [56]2 years ago
4 0

Answer:

Explanation:

Given

mass of tree stump is M

mass bullet is m

velocity of bullet is v

Conserving momentum for bullet and tree stump

Initial Momentum P_i=mv

Suppose v_0 is the velocity of the system

Final Momentum P_f=(M+m)v_0

Initial momentum =Final Momentum

mv=(m+M)v_0

v_0=\frac{mv}{m+M}

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2 years ago
Explain how wind acts as an agent of erosion and deposition
Maksim231197 [3]
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3 years ago
Question 8 of 10
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2 years ago
Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.
ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

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3 years ago
An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i
Svetllana [295]

Answer:

v \approx 9.312\,\frac{m}{s}

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}

K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}

\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot s\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cdot s \cos \theta

\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)

v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}

v \approx 9.312\,\frac{m}{s}

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3 years ago
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