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jeyben [28]
2 years ago
13

Calculate the momentum pelephant of a 2140 kg elephant charging a hunter at a speed of 7.63 m/s . pelephant= kg⋅m/s Compare the

elephant's momentum with the momentum pdart of a 0.0415 kg tranquilizer dart fired at a speed of 619 m/s . pelephant=pdart pelephant>pdart More information is needed to determine how the momentums compare. pelephant
Physics
1 answer:
Zepler [3.9K]2 years ago
7 0

Answer:

Momentum of elephant is greater than momentum of tranquilizer dart

Explanation:

m_e = Mass of elephant = 2140 kg

m_d = Mass of tranquilizer dart = 0.0415 kg

v_e = Velocity of elephant = 7.63 m/s

v_e = Velocity of tranquilizer dart = 619 m/s

Momentum of elephant

p_e=m_ev_e\\\Rightarrow p_e=2140\times 7.63\\\Rightarrow p_e=16328.2\ N.s

Momentum of tranquilizer dart

p_d=m_dv_d\\\Rightarrow p_d=0.0415\times 619\\\Rightarrow p_d=25.6885\ N.s

So, p_e>p_d

Momentum of elephant is greater than momentum of tranquilizer dart

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In a circus act, a 70 kg clown is shot from a cannon with an initial velocity of 17 m/s at some unknown angle above the horizont
EleoNora [17]

Answer:

K_{2}=7302.4J

Explanation:

Given the initial velocity of the clown, his mass and final height we can calculate the final kinetic energy using the <em><u>conservation of total mechanical energy</u></em>

K_{1}+U_{1}=K_{2}+U_{2}

K_{2}=K_{1}+U_{1}-U_{2}

K_{2}=\frac{1}{2}mv_{1}^{2}+mgh_{1}-mgh_{2}

Since h_{1}=0

K_{2}=\frac{1}{2}mv_{1}^{2}-mgh_{2}

K_{2}=\frac{1}{2}(70kg)(17m/s)^2-(70kg)(9.8m/s^2)(4.1m)=7302.4J

K_{2}=7302.4J

7 0
3 years ago
A man is traveling from the back of a boat to the front of the boat at 2.0 m/s while the boat itself is traveling at 12.0 m/s to
Hatshy [7]

Based on the relative velocity of the man with respect to the boat and the dock:

  • Distance covered in 4.0 seconds relative to the boat  = 8 m
  • Distance covered in 4.0 seconds relative to the dock = 56 m

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of a body relative to another body which serves as a reference point.

Relative velocity is a vector.

Considering the velocity of the man and the boat:

The relative velocity of the man with respect to the boat = 2.0 m/s

Distance covered in 4.0 seconds relative to the boat = 2.0 m/s * 4.0 s

Distance moved = 8 m

Relative velocity of the man with respect to the dock = 12 + 2 = 14 m/s

Distance covered in 4.0 seconds relative to the dock = 14.0 m/s * 4.0 s

Distance moved = 56 m

In conclusion, the relative velocity is velocity with respect to a reference point.

Learn more about relative velocity at: brainly.com/question/24337516

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1 year ago
On January 22, 1943, the temperature in Spearfish, South Dakota, rose from −4°F to 45°F in just 2 minutes. What was the temperat
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The answer is the first one
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2 years ago
Friction pulls directly against the direction of motion (at 180) of a sled, and does -55.2 J of work while the sled moves 8.98 m
zvonat [6]

The magnitude of the force of friction is 6.1 N

Explanation:

The work done by a force when moving an object is given by the equation:

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

W = -55.2 J (work done by the force of friction)

d = 8.98 m (displacement of the sled)

\theta=180^{\circ} (because the force of friction acts opposite to the direction of motion)

By solving the equation for F, we find the magnitude of the force of friction on the sled:

F=\frac{W}{d cos \theta}=\frac{-55.2}{(8.98)(cos 180^{\circ})}=6.1 N

Learn more about work:

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