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9966 [12]
3 years ago
15

Which of the following are examples of energy being conserved? Choose all that apply A light bulb constantly changes electrical

energy into light energy. A plant collects sunlight to form glucose, storing the energy as potential energy. A car uses its breaks to stop suddenly, causing the tires to heat up. Your friend proposes an idea for a fan that doesn't require electricity or a fuel source.
Physics
2 answers:
Y_Kistochka [10]3 years ago
5 0

The only one on that list that DOESN'T involve conservation of energy is your friend's cockamamie idea.  

-- A light bulb constantly changes electrical energy into light energy.  The heat energy and light energy coming out of the bulb add up to exactly the amount of electrical energy that it consumes.  Energy is conserved.

-- A plant collects sunlight to form glucose, storing the energy as potential energy.  The plant stores CHEMICAL energy in the glucose ... the same amount as the energy it absorbed from the sunlight.  Energy is conserved.

-- A car uses its breaks to stop suddenly, causing the tires to heat up.  The car's brakes (sp!) convert some of the car's kinetic energy ... slowing or stopping the car ... into that same amount of heat energy.  Energy is conserved.

-- Your friend proposes an idea for a fan that doesn't require electricity or a fuel source.  Well first of all, the heat that your friend's brain produced in the process of THINKING up the idea WAS the same as the amount of chemical energy from the food he ate that he used to operate his brain.  Energy is conserved in THAT part of the process.  

And that's why his idea will never work.  He's telling you that his fan will turn itself (kinetic energy) and move some air (more kinetic energy), but you won't have to put any energy INTO it.  So he claims that new energy, that never existed before, will be created inside the fan, and if you run the fan inside a closed, insulated room, the amount of energy inside the room will increase.

IF there was any way that this could work, your friend could just run his fan all day, SELL the energy that it manufactured, and be the most fabulously wealthy individual who ever lived.

It's never gonna happen.  His idea can't work, because energy is not conserved in it.

Degger [83]3 years ago
4 0
A plant collects sunlight to form glucose, and your friend proposes an idea for a fan. Conserved = saving
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The answers to the problem are as follows:

MA= 5 

IMA= input distance/ output distance, 5 

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x is vertical and y is horizontal

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Mass is a measure of weight. True False
8_murik_8 [283]

Answer: False

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2 years ago
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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1
xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

8 0
2 years ago
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