Artificial satellites in space can help you find locations on<br> Earth. True or false?

Amotor supplied by 240V requires 12A to lift a 2000 lb at a rate of 25 ft/min, me power input to the motor is-

tensa zangetsu [6.8K]

Answer:

Power input, P = 2880 watts

Explanation:

It is given that,

Voltage of the motor, V = 240 V

Current required, I = 12 A

Weight lifted, W = 2000 lb

It is lifting at a speed of 25 ft/min. We need to find the power input to the motor. The product of current and voltage is called power input of the motor.

$P=I\times V$

$P=12\ A\times 240\ V$

P = 2880 watts

So, the power input of the motor is 2880 watts. Hence, this is the required solution.

4 0

4 years ago

A hand-held video game is powered by batteries. After playing the game for several minutes, a student notices that the game feel

scoundrel [369]

I would say your answer is B- Some of the chemical energy from the batteries is converted into heat energy.

7 0

3 years ago

Is it true muscle mass affects someones flexibility

guapka [62]

It is very very true

8 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$