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2 years ago

A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.

2 answers:

7 0

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

The body is gaining the speed, which means there is a change in kinetic energy.
The change in kinetic energy is equal to the work done.
The friction force is the product of coefficient of the friction and normal force.
The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

Zarrin [17]2 years ago

4 0

Answer:

Explanation:

The normal force will be perpendicular to the motion. A zero-work situation is one where "a force everywhere perpendicular to the motion does no work" so according to this principle, no work is done.

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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.

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3 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

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3 years ago

With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

hichkok12 [17]

(2.0 x 10² N/Coul) x (-1.6 x 10⁻¹⁹ Coul) = -3.2 x 10⁻¹⁷ N

The magnitude is 3.2 x 10⁻¹⁷ Newton.

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3 years ago

What is the source of all electromagnetic waves?

kherson [118]

Answer:

Accelerating charges.

Explanation:

Electromagnetic waves are waves produced by the vibration of both electrical and magnetic fields.

This interaction produces an energy source that does not require any medium to propagate.

To produce electromagnetic waves, electric and magnetic fields must be vibrating.

An electric charge produced when vibrating under voltage will produce electromagnetic waves. This is the same for all sources of these waves.

The sun produces electromagnetic waves. A lot of human activities also does this.

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3 years ago

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lukranit [14]

Answer:

an increase in gasses that absorb heat

Explanation:

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Greenhouse effect leads to increase in the temperature of the earth, melting of polar ice caps and possibly flooding due to a rise in sea levels.

Greenhouse gases act as glass in a greenhouse. They allow heat to pass through onto the earth surface but trap the heat and prevent it from being radiated outwards back to space. Thereby increasing the surface temperature of the earth.

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4 years ago

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