All categories

3 years ago

Select the correct answer from each drop-down menu.

Physics

2 answers:

Thepotemich [5.8K]3 years ago

6 0

Answer:

Gravitational potential energy is converted into kinetic energy

Explanation:

During an avalanche, the gravitational potential

energy of the snow on the mountain is converted into

kinetic energy as the snow cascades down.

The potential energy stored by the snow collected high in the mountain under the gravitational field created by our Earth, breaks loose and as it comes down acquiring velocity, it is converted into kinetic energy due to its accelerated motion

Aleks [24]3 years ago

3 0

Answer:

1. Potential energy.

2.kinetic energy.

You might be interested in

If you lived by the sea, what effect of the moon would you see? Describe the effect.

dolphi86 [110]

Answer:

the changes in waves

Explanation:

the moon has its own gravitational pull thus making waves and the rising tides

8 0

3 years ago

A ball is rolling down a hill. Wich action would slow the ball down?

Feliz [49]

Answer:

Friction Force

hope this helped :)

Explanation:

4 0

2 years ago

Read 2 more answers

how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy

Blababa [14]

This question is probably referring to heat energy transferring from the car to its surroundings.

4 0

3 years ago

A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10

Alchen [17]

Answer:

v = 895 m/s

Explanation:

Time period is given as 39 Earth Days

$T = 39 days \times 24 hr \times 3600 s$

$T = 3369600 s$

now the radius of the orbit is given as

$r = 4.8 \times 10^8 m$

so the total path length is given as

$L = 2 \pi r$

$L = 2\pi (4.8 \times 10^8)$

$L = 3.015 \times 10^9$

now the speed will be given as

$v = \frac{L}{T}$

$v = \frac{3.015 \times 10^9}{3369600}$

$v = 895 m/s$

7 0

3 years ago

What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300

Vinvika [58]

The efficiency of an ideal Carnot heat engine can be written as:
$\eta = 1- \frac{T_{cold}}{T_{hot}}$
where
$T_{cold}$ is the temperature of the cold region
$T_{hot}$ is the temperature of the hot region

For the engine in our problem, we have $T_{cold}=300 K$ and $T_{hot}=400 K$ , so the efficiency is
$\eta= 1 - \frac{300 K}{400 K}=0.25$

4 0

3 years ago