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zhuklara [117]
3 years ago
5

What are the limitations of using solar panels?

Physics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

The amount of the sun energy that could be collected.

Explanation:

Some limitations are, the amount of the sun's energy that could be collected the radiation of sun is nearly fixed. The place where we can put the solar panels are also limited.

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When you are ice skating, to get started, you push your skate backwards on the ice and, as a result, begin to move forward. Whic
Setler79 [48]
I would say the third... force. 

3 0
3 years ago
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A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the bl
ss7ja [257]
Kinetic energy =1/2 mv^2 

<span>m=2ke/v^2 </span>

<span>m=2(34)/3.6^2 </span>

<span>m=5.24 </span>

<span>force normal = mg </span>
<span>=5.24 x 9.8 </span>
<span>force normal = 51.4N

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.


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3 years ago
Which force is greater the earth’s pull on the moon or the moon’s pull on the earth
Vitek1552 [10]

Answer:

The earth's pull on the moon

Explanation:

Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth.

7 0
3 years ago
A protein molecule in an electrophoresis gel has a negative charge.The exact charge depends on the pH of the solution, but 30 ex
Ad libitum [116K]

Answer:

The magnitude of the electric force on a protein with this charge is 7.2\times10^{-15}\ N

Explanation:

Given that,

Electric field = 1500 N/C

Charge = 30 e

We need to calculate the magnitude of the electric force on a protein with this charge

Using formula of electrostatic force

F=Eq

Where, F = force

E = electric field

q = charge

Put the value into the formula

F=1500\times30\times1.6\times10^{-19}

F=7.2\times10^{-15}\ N

Hence, The magnitude of the electric force on a protein with this charge is 7.2\times10^{-15}\ N

6 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
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