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Serhud [2]
3 years ago
10

A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engin

e. How many ropes are supporting the engine
Physics
1 answer:
Sidana [21]3 years ago
8 0

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

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luda_lava [24]

Answer:

I think it's 0 N 3rd choice

4 0
3 years ago
Match these terms with the correct examples.
posledela
1. liquid solution to a. oceans
2. gaseous solution to b. clouds
Not sure about 3 and 4.
3 might be oxygen but I think that's 5. element.

Hope this helps, not sure about water and air though.
4 0
3 years ago
Read 2 more answers
hergy A Sprinter runs at a a forward velocity of 10,9 m/s. If the sprinter has a mass of 72.5 kg, What is the Sprinter's Kinetic
Nina [5.8K]

Answer:

4306.8625 J or 344549/80J (fraction form)

Explanation:

kinetic energy= \frac{1}{2\\}mv²

= \frac{1}{2}×72.5×10.9²

= 4306.8625 J  or 344549/80 J

7 0
1 year ago
Lightweight, vertically suspended spiral spring with a spring constant of 8.6 N / m is fitted with 64 g weight. The weight shall
alexandr1967 [171]

Answer:

Explanation:

Given that

Force constant k=8.6N/m

Weight =64g=64/1000=0.064kg

Extension is 45mm=45/1000= 0.045m

It will have it highest spend when the Potential energy is zero

Therefore energy in spring =change in kinetic energy

Ux=∆K.e

½ke² = ½mVf² — ½mVi²

Initial velocity is 0, Vi=0m/s

½ke² = ½mVf²

½ ×8.6 × 0.045² = ½ ×0.064 ×Vf²

0.0087075 = 0.032 Vf²

Then, Vf² = 0.0087075/0.032

Vf² = 0.2721

Vf=√0.2721

Vf= 0.522m/s

The time it will have this maximum velocity?

Using equation of motion

Vf= Vi + gr

0.522= 0+9.81t

t=0.522/9.81

t= 0.0532sec

t= 53.2 milliseconds

5 0
3 years ago
To determine the pressure in a fluid at a given depth with the air-filled cartesian diver, we can employ Boyle's law, which stat
aniked [119]

Answer:

The pressure at this depth is 1.235\cdot P_{atm}.

Explanation:

According to the statement, the uncompressed fluid stands at atmospheric pressure. By Boyle's Law we have the following expression:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

V_{1}, V_{2} - Initial and final volume.

P_{1}, P_{2} - Initial and final pressure.

If we know that V_{2} = 0.81\cdot V_{1}, then the pressure ratio is:

\frac{P_{2}}{P_{1}} = 1.235

If P_{1} = P_{atm}, then the final pressure of the gas is:

P_{2} = 1.235\cdot P_{atm}

The pressure at this depth is 1.235\cdot P_{atm}.

6 0
3 years ago
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