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Oksana_A [137]
3 years ago
12

Over the last several decades scientists have considered the problem of nonrenewable natural resources, such as fossil fuels. Hu

mans are using fossil fuels at a rapid rate and scientists believe that one day, we will run out of usable fossil fuels. To solve this problem, research has been conducted on alternative energy sources. One alternative energy source is wind energy. Although wind energy is a good solution to this problem, it is not the perfect solution.
Which statement describes a drawback to wind energy?
A) Wind energy is a clean source of energy.
B) Good locations for turbines are limited.
C) Wind energy is inexpensive to set up and use.
D) Wind energy plants do not involve the use of hazardous chemicals.
El
Physics
1 answer:
sesenic [268]3 years ago
8 0

The statement "Good locations for turbines are limited" describes a drawback to wind energy.

Answer: Option B

<u>Explanation:</u>

Wind energy is one of the most useful and efficient renewable energy sources. But nothing is ideal in this universe and the same thing applies for wind energy also. The generation of electricity from wind energy requires setting up of turbines.

And these turbines can be set up in plane areas which is free from any disturbance except wind flow. In open area and flat plane surface only the turbines can rotate freely with the effect of wind.

But regions where the wind flow is minimum due to snow formation like the northern region of earth, the turbines cannot be set up there. So the locations for setting up of turbines are limited for good outcome in wind energy. This is one of the drawback of wind energy.

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A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

P = VI

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

V = IR

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

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