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3 years ago

What do you need to build a working electric motor?

Physics

1 answer:

FrozenT [24]3 years ago

5 0

Answer: D a source of electricity

Explanation: I have done that question please tell me if it is correct

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Oduvanchick [21]

Answer:

matter: A

data: E

variable: C

Controlled: D

Physical: B

Explanation:

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2 years ago

Two large, plastic tubs were filled with soil The soil was shaped to create a mound in each tub. The starting height of each mou

shutvik [7]

Answer:

Explanation:

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2 years ago

Lucas left a metal bowl and a wooden bowl outside overnight. The next morning, he picked up the bowls to bring them inside. He n

kipiarov [429]

Answer: C. Metal transfers heat away from the skin by conduction, creating the sensation of coolness.

Explanation: The skin releases heat into the metal bowl since there is a difference in temperature between the two objects. So heat is taken away from the hand abd transfers into the metal bowl by conduction creating a cooler sensation.

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3 years ago

How do you find the force of gravity?

devlian [24]

1Define the equation for the force of gravity that attracts an object, Fgrav = (Gm1m2)/d2
2. Use the proper metric units.
3. Determine the mass of the object in question.
4. Measure the distance between the two objects
5. Solve the equation

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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:

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:

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

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 %.

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

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