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andrey2020 [161]
4 years ago
8

When energy is changed from one form to another, some of the energy is lost to the surroundings as waste thermal energy. The mor

e steps needed to arrive at the desired energy or the more types of energy that are generated, the more energy is wasted.
Which object would likely have the least amount of wasted energy?

A) an electric drill
B) a solar cell
C) a car engine
D) a hydroelectric dam
Physics
2 answers:
Alex4 years ago
8 0

Answer: The correct answer is solar cell.

Explanation:

When energy is changed from one form to another, some of the energy is lost to the surroundings as waste thermal energy.

From the given options,

Electric drill: Electric drill converts the electrical energy into mechanical energy. When the drill break the the object then the mechanical energy is converted into the heat energy.

Car engine: Car engine converts the chemical energy into the thermal energy. This thermal energy gets converted into the mechanical energy. Then, this mechanical energy is used to accelerate the vehicle.

Hydroelectric dam: Here, the potential energy stored in the water at some height is converted into mechanical energy. Then, this mechanical energy is converted into the kinetic energy. Then, finally, this kinetic energy is converted into electrical energy.

In above options, there is a dissipation of energy while one form of energy gets converted into another.

The more steps needed to arrive at the desired energy or the more types of energy that are generated, the more energy is wasted.

Solar cell: The solar cell converts the solar energy into electrical energy.

Therefore, a solar cell have the least amount of wasted energy.

Taya2010 [7]4 years ago
7 0
The answer is B Solar cell just took the test
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Romashka [77]

Answer:

Average Speed = 6.37 m/s

Explanation:

The average speed is simply given by the following formula:

Average Speed = Total Distance Traveled/Total Time Spent

here,

Total Time Spent = 1.1 min + 1.5 min = (2.6 min)(60 s/min) = 156 s

Now, for total distance, we have to calculate the distance traveled on tortoise and distance traveled while flying, separately. Therefore,

Distance Traveled on Tortoise = (Time spent on Tortoise)(Speed of Tortoise)

Distance Traveled on Tortoise = (1.1 min)(60 s/min)(0.06 m/s) = 3.96 m

Similarly,

Flying Distance = (Flying Time)(Flying Speed) = (1.5 min)(60 s/min)(11 m/s)

Flying Distance =  990 m

Since, total distance is the sum of both distances, therefore,

Total Distance = 3.96 m + 990 m = 993.96 m

Now, using the values in equation of average speed, we get:

Average Speed = 993.96 m/156 s

<u>Average Speed = 6.37 m/s</u>

4 0
3 years ago
What unit is used to measure the amount of energy used?
NNADVOKAT [17]

Answer: Joule

Explanation:

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3 years ago
How much heat in joules is required to heat a 50 g sample of aluminum from 71 ∘f to 142 ∘f?
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3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

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