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Leya [2.2K]
2 years ago
10

ASAP

Physics
1 answer:
iris [78.8K]2 years ago
5 0

Answer:

A) Energy is dissipated into heat and sound energy due to Friction

B) The energy goes into heat and sound energy due to friction again, otherwise the cart would accelerate due to an unbalanced force. Therefore, we know there's friction, and the friction causes energy loss.

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A person jumps from the roof of a house 3.1-m high. When he strikes the ground below, he bends his knees so that his torso decel
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Part A

Free fall motion
h = 3.1 m

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That is the only part in the question.
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Water raises a boat once every 3.0 seconds. What is the frequency (f) of the waves passing the boat?
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Explanation:

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If a change is made to a system in equilibrium, the equilibrium will shift to oppose the change.
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Read 2 more answers
Assume that Parker Company will receive SF200,000 in 360 days. Assume the following interest rates:
Anna35 [415]

Answer:

b.  $96,914

Explanation:

360-day borrowing rate = 5%

spot rate = 0.48

360-day deposit rate  = 6%

Borrow at the rate of 5% to get

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Convert at the spot rate of $0.48 to get

190,476.19*0.48 = $91,428.57

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91,428.57/1.06 = 96,914.28

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7 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
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