How does energy move predictably between a lien water in the air above it

2. physical quantities which do not depend on any physical quantities for their measurements are known as

seropon [69]

Answer:

Fundamental quantities

5 0

2 years ago

12. your friend with great excitement tells you about his newest idea to solve the energy crisis: he wants to use an electromoto

mixas84 [53]

I would tell him, in the kindest, most gentle way I could manage,
to fahgeddaboudit.

The total amount of energy doesn't change. Energy is never created,
and it never disappears. If you have some energy, then it had to come
from somewhere, and if you used some energy, then it had to go
somewhere.

You can never get more energy out of the electromotor than you put into it,
and in the real world, you can't even get THAT much out, because some
of it is always used on the way through.

Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
"Free Energy" on the internet, relax, and enjoy the show. They are all
fakes. They may not all be intentionally meant to fool you, but they are
all impossible.

4 0

3 years ago

Read 2 more answers

A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the

Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

$a_c =w^2r = \frac{v^2}{r}$

in this case we know the speed of the runner

$v =8.00\ m/s$

The radius "r" will be the distance from the runner to the center of the track

$r = 28.2\ m$

$a_c = \frac{8^2}{28.2}\ m/s^2$

$a_c = 2.27\ m/s^2$

The answer is the last option

3 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$