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ser-zykov [4K]
3 years ago
9

In order for work to be done, what must be true?

Physics
1 answer:
Amiraneli [1.4K]3 years ago
5 0

Answer:

option c

Explanation:

Work is said to be done when a force moves a body over a distance, in the direction of force applied.

Work done = force x distance (in the direction of the force)

Therefore work = force x displacement

As displacement is distance in a specific or specified direction.

I hope this was helpful, Please mark as brainliest

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Why isnt geothermal energy used more often​
Dafna1 [17]
Risk of return on investment is higher than other forms of energy generation.
6 0
4 years ago
The Old Faithful geyser found in Yellowstone Park is an example of _____.
katen-ka-za [31]

Answer:

cone geyser

Explanation:

The Old Faithful geyser is the oldest discovered geyser in the Yellowstone national park. The eruptions of the geyser are particularly predictable. It is a cone type geyser.

Cone geysers generally have a spout through which the water ejects out. When super heated water in the tube then the water starts to boil and form bubbles of steam, after this process the eruption takes place.

5 0
3 years ago
1. Give three examples, from the lab, where potential energy was converted to kinetic energy: ​
Lubov Fominskaja [6]

Answer:

A book on a table before it falls.

A yoyo before it is released.

A raised weight.

Explanation:

These are all examples of potential energy. So I hope you can find something that is comparable from the lab.

3 0
3 years ago
How long would it take a machine to do 5.000
Gala2k [10]

Answer:

a

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

6 0
3 years ago
In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s
Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle =  35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ   with x axis    .............1

m1v3sin35 = m2v4sinθ                   with y axis  .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ  = 3.38                                            .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ  

v4 sinθ = 1.95                                              ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0
3 years ago
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