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earnstyle [38]
3 years ago
7

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

haft by a cable connected to a winch. The shaft is inclined at 34.0° above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed.
Physics
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

11.714 kW

Explanation:

Here is the complete question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0∘ above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed. What power must the winch motor provide when the car is moving at constant speed?

Solution

Since the loaded ore car moves along the mine shaft at an angle of θ = 34° to the horizontal, if F is the force exerted on the cable, then the net force on the laoded ore car is F - mgsinθ = ma where  mgsinθ = component of the car's weight along the incline, m = mass of loaded ore car = 950 kg and a = acceleration

F = m(a + gsinθ)  

When the car is moving at constant speed, a = 0

So F = m(a + gsinθ) = F = 950(0 + 9.8sin34) = 5206.1 N

Since it continues at a constant speed of v = 2.25 m/s, the power of the winch motor is P = Fv = 5206.1 N × 2.25 m/s = 11713.7 W = 11.714 kW

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Answer:

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Part B  Mica- mylar- paper- quartz

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Another magnitude that we must take into account that the maximum working voltage, the greater the safer is the capacitor

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Part A the answer is the dielectric constant.

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Mica. The best ever

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3 years ago
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Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

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3 years ago
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A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s
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Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

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Apply the following equation to determine the speed of the insect which is the source;

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Therefore, the speed the bat is gaining on its prey is 0.03m/s

8 0
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