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Lady_Fox [76]
3 years ago
15

Nate the Skate was an avid physics student whose main non-physics interest in life was high-speed skateboarding. In particular,

Nate would often don a protective suit of Bounce-Tex, which he invented, and after working up a high speed on his skateboard, would collide with some object. In this way, he got a gut feel for the physical properties of collisions and succeeded in combining his two passions. On one occasion, the Skate, with a mass of 129 kg, including his armor, hurled himself against an 831 kg stationary statue of Isaac Newton in a perfectly elastic linear collision. As a result, Isaac started moving at 1.29 m/s and Nate bounced backward. What were Nate's speeds immediately before and after the collision? Ignore friction with the ground. Choose the correct option;
(a) before collision: 4.8m/sAfter Collision: 1.29m/s
(b) before collision: 2.8m/sAfter Collision: 2.29m/s
(c) before collision: 3.8m/sAfter Collision: 3.29m/s
(d) before collision: 5.8m/sAfter Collision: 1.29m/s
Physics
1 answer:
Ivenika [448]3 years ago
4 0

Answer:

v_{1i} = 4.8 m/s

v_{1f} = 3.51 m/s

Explanation:

Since the collision is perfectly elastic collision so here we can use momentum conservation

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

so we will have

129 v_{1i} + 0 = 129(-v_{1f}) + 831(1.29)

v_{1i} + v_{1f} = 8.31

also by elastic collision property we will have

1.29 + v_{1f} = v_{1i}

now from above two equations we have

v_{1i} = 4.8 m/s

v_{1f} = 3.51 m/s

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We have that the sea level pressure for Leh area is 1150mb mathematically given as

Ps= 1150 mb

<h3> Sea level pressure</h3>

Question Parameters:

Ladakh is 800 mb.

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2 years ago
The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo
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Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

<u>Explanation:</u>

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen.  At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution.  In the event, that the net torque is zero, at that point angular momentum is steady or saved.  

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework.  Be that as it may, we likewise double the moment of inertia. Since L=I \times \omega, the turning rate of the two-puck framework must stay unaltered.

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3 years ago
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
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Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

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The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

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The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

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