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Leya [2.2K]
3 years ago
7

Explain how kinetic energy transforms into potential energy. Give at least 2 examples.

Physics
1 answer:
Lena [83]3 years ago
5 0
When a ball is tossed into the air and lands on top of a hill. (Kinetic when it's flying in the air. Potential when it's at the top about to roll down)

When a skater goes down a ramp and goes back up but waits to go back down
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Imagine how mercury might be different if it had the same mass as earth. Explain.
maxonik [38]
I believe if it were heavier with more mass, then the sun would pull it in and there would be no mercury. It might also be hotter.
4 0
3 years ago
Please Answer these quick!!!! I WILL GIVE BRAINLIEST<br> Do as many as you can!
shepuryov [24]

Answer:

i

said

1 a

2 b

3 a

\idonotcare.net

Explanation:

6 0
3 years ago
PLS HELP!!! A 1200-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationa
Fittoniya [83]

Answer:

7200 kg.m/s

Explanation:

According the law of conservation of linear momentum, the sum of momentum before and after collision are equal.

Using this principle, the sum of initial momentum will be given as p=mv where p is momentum, m is mass and v is velocity

Initial momentum

Mass of whale*initial velocity of whale + mass of seal*initial seal velocity

Since the seal is initially stationary, its velocity is zero. By substitution and taking right direction as positive

Initial momentum will be

1200*6+(280*0)=7200 kg.m/s

Since both initial and final momentum should be equal, hence the final momentum will also be 7200 kg.m/s

7 0
3 years ago
The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values
Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r²    ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m    = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0
3 years ago
Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.
fredd [130]

Energy to lift something =

               (mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm  =  5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

            =  2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

7 0
2 years ago
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