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Bumek [7]
3 years ago
11

A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down

the ramp?
Physics
1 answer:
Len [333]3 years ago
4 0

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

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fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

Xc = \frac{1}{2\pi fC}

where,

Xc\\ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

Xc =?

Xc' = \frac{1}{2\pi f'C}

      = \frac{1}{2\pi 2f C}

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Xc' = \frac{1}{2} Xc

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

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3 0
2 years ago
A woman of mass 50 kg is swimming with a velocity of 1.6 m/s. If she stops stroking and glides to a stop in the water, what is t
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Answer:

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Explanation:

<u>Given the following data;</u>

Mass = 50kg

Initial velocity = 1.6m/s

Since she glides to a stop, her final velocity equals to zero (0).

Now, we would find the change in velocity.

Change \; in \; velocity = final \; velocity - initial \; velocity

Substituting into the equation above;

Change in velocity = 0 - 1.6 = 1.6m/s

Impulse \; of \; force = mass * change \; in \; velocity

Substituting into the equation, we have;

Impulse \; of \; force = 50 * -1.6

<em>Impulse of force = -80 Ns</em>

<em>Therefore, the impulse of the force that stops her is -80 Newton-seconds and it has a negative value because it is working in an opposite direction, thus, bringing her to a stop. </em>

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3 years ago
A 100kg bag of sand has a weight on 100 N. When dropped it's acceleration is what?​
Vesnalui [34]

Answer:

9.8m/s

Explanation:

acceleration due to gravity is independent of mass

7 0
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The atoms in a sidewalk on a hot summer day are vibrating _____ the atoms in the sidewalk on a freezing day in winter.
Karolina [17]

Answer:

(A) more rapidly than

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With higher temperatures, object's molecules (and atoms) have higher kinetic energy which is due to faster "jiggling" (vibrations). On a hot day these vibrations in the material the sidewalk is made of are more rapid than on a cold day, just as their temperatures differ.

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