rocks move trough the rock cycle through uplift,folding,and faulting cased by movements of the earth´s_______

Shkiper50 [21]

Edit

In physics, power is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time. Having no direction, it is a scalarquantity. In the International System of Units, the unit of power is the joule per second (J/s), known as the watt in honour of James Watt, the eighteenth-century developer of the condenser steam engine. Another common and traditional measure is horsepower (comparing to the power of a horse). Being the rate of work, the equation for power can be written:

Power

Common symbols

Derivations from

other quantities

P = E/t

P = F·v

P = V·I

P = T·ω

As a physical concept, power requires both a change in the physical system and a specified time in which the change occurs. This is distinct from the concept of work, which is only measured in terms of a net change in the state of the physical system. The same amount of work is done when carrying a load up a flight of stairs whether the person carrying it walks or runs, but more power is needed for running because the work is done in a shorter amount of time.

8 0

4 years ago

The humidity you feel in the southern states of the U.S. is _____. part of the water cycle water vapor caused by the Sun's heat

Likurg_2 [28]

Hey there!

The answer is :
The <span>humidity you feel in the southern states of the U.S. is water vapor
water vapor is between a gas and liquid so that is what gives a humid feel in the hot sun

Hopes this helps u :D</span>

4 0

3 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$