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3 years ago

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a) If a proton moved from a location with a 5.0 V potential to a location with 7.5 V potential, would its potential energy incre

ase or decrease? Explain.b) If an electron moved from a location with 7.5 V potential to a location with a 5.0 V potential, would its potential energy increase or decrease? Explain

1 answer:

tekilochka [14]3 years ago

3 0

Answer:

A. The potential energy of the positively charge proton from 5v to 7.5v will increase because in an electric field work is done on the proton to move it to a point of higher potential

B. when an electron is moved from 7.5v to 5v it loses potential energy due to the charges lost so as to bring it to a point of lower potential

Explanation:

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A nuclear power plant operates at 79 percent of its maximum theoretical (Carnot) efficiency between temperatures of 700° and 330

ololo11 [35]

Answer:

Explanation:

T₁ = 700 + 273 = 973 k

T₂ = 330 + 273 = 603 k

Theoretical efficiency = T₁ - T₂ / T₁

= (973 - 603) / 973

= .38 OR 38%

Operating efficiency = .79 x 38

= 30.02 %

Heat input Q₁ , Heat output to sink Q₂ , conversion into power = Q₁ - Q₂

given Q₁ - Q₂ = 1.3 x 10⁹ W

efficiency = Q₁ - Q₂ / Q₁

Q₁ - Q₂ / Q₁ = 30.02 / 100

100Q₁ - 100Q₂ = 30.02Q₁

69.98 Q₁ = 100Q₂

Q₁ = 1.429 Q₂

Putting this in the relation

Q₁ - Q₂ = 1.3 x 10⁹ W

1.429Q₂ - Q₂ = 1.3 x 10⁹ W

.429Q₂ = 1.3 x 10⁹

Q₂ = 3.03 x 10⁹W

= 3.03 GW.

6 0

3 years ago

A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is

Arada [10]

Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image

$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$

1 / p = 1 / 2.2 - 1/3.29

1 / p = 0.15059

p = 6.64 cm

therefore the farthest distance from the object is 6.64 c

3 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

This is the new external F force that we are looking for:

$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

and since: μs* $mg=41.6N$

the new F force would be:

$F=3mg$ *μs $=41.6*3=124.8N$

4 0

3 years ago

calculate the percentage increase in speed of the cyclist when the power output changes from 200W to 300W

Likurg_2 [28]

Answer:

50%

Explanation:

That would be the amount

4 0

3 years ago

You have 0.5 l of air at a pressure of 203 kpa and -70°c in a rigid, sealed contaainer. what is the absolute tempertaure of the

ANTONII [103]

From the information given and if the question is complete then;
Absolute temperature is the temperature in Kelvin
To convert degree Celsius to kelvin we normally add 273
that is Kelvin = deg Celsius + 273
Thus since we have been given that the air was at -70 degrees celcius;
then; - 70° C + 273 = 203 K
Therefore; the absolute temperature is 203 K

5 0

3 years ago