What is the process in which a substance changes from a solid into a liquid?

While an emf source supplies energy to a circuit, that energy is dissipated when the current passes through resistance. In deali

fredd [130]

The question is incomplete, the options are;

RI^2

I^2/R

R/I^2

R/V^2

RV^2

V^2/R

VI

VIR

Select all that apply

Answer:

P=RI^2

P=V^2/R

P=VI

Explanation:

Power is the rate at which energy is changing in a circuit. It is shown by the formulas outlined above from the group of answer choices. Since the current (I), voltage (V), and resistance (R) were mentioned in the question, any of three three formulas could be used to obtain the power drawn by the conductor.

7 0

3 years ago

A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?

givi [52]

I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.

If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.

For the wave you described . . .

   Frequency = (speed) / (wavelength)

   = (6 m/s) / (12 m)

   = 0.5 / sec

   =    0.5 Hz .

(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)

5 0

3 years ago

Read 2 more answers

A bowling (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform

cestrela7 [59]

Hope this helps you!

7 0

3 years ago

microwave ovens rotate at a rate of about 6.3 rev/min. what is this in revolutions per second? (you do not need to enter any uni

levacccp [35]

The microwave ovens rotate at a rate of about 0.105 rev/s.

The microwave rotation is the number of revolutions in a unit of time. To change the unit for angular velocity, assume that the quantity is multiplied by the unit it has. Then change to the desired units. The angular velocity is denoted by ω and has a magnitude of 6.3 rev/min.

ω = 6.3 rev/min

$\omega = 6.3 \times \frac{1 \: rev}{1 \: min}$

1 minute = 60 seconds
The revolution unit didn't change

$\omega = 6.3 \times \frac{1 \: rev}{60 \: seconds}$

$\omega = \frac{6.3 \: rev}{60 \: seconds}$

ω = 0.105 rev/s

Learn more about Angular velocity here: brainly.com/question/29344944

#SPJ4

7 0

1 year ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: