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Murljashka [212]
3 years ago
14

Which one of the two potential differences emf or terminal voltage will be greater in magnitude?Why?​

Physics
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

True, yes terminal voltage of a battery can never be greater than the emf of the battery. Because due to internal resistance of the current terminal

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A book is sitting on a table. Jackie is pushing the book with a force of 9 N to the right. If the force of friction is 4 N to th
Free_Kalibri [48]

Answer:

5 N right

Explanation:

Fx = 9N-4N

Fx = 5N

Since we can define the x and y axis. We have x to the right as positive.

8 0
2 years ago
Conveyor belts are often used to move packages around warehouses. The conveyor shown below moves packages at a steady 4.0 m/s. A
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Answer:

0 j

Explanation:

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3 years ago
Plz help me asap !!!!!!!!!!
nlexa [21]

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B- 65km

C- 2.8km/h

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3 years ago
If a specimen was being viewed using a 20x objective lens and 10x ocular lens, what would be the total magnification
Paraphin [41]

Answer:

As Per Given Information

20x objective lens was used by specimen

10x ocular lens was also used by him.

we have to find the total magnification.

For calculating the total magnification we 'll simply do multiplication

Total Magnification = 20x × 10x

Total Magnification = 200x

So , the total magnification will be 200x .

6 0
2 years ago
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0
3 years ago
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