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saw5 [17]
3 years ago
12

When does a object exerts an action force

Physics
1 answer:
nlexa [21]3 years ago
8 0

Answer: According to Newton's third law of motion, whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. These two forces are called action and reaction forces.

Explanation:

No cap i know

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Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for
klemol [59]

The value of parameter C for the function in the figure is 2.

<h3>What is amplitude of a wave?</h3>

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(x - C)

where;

  • A is amplitude of the wave
  • C is phase difference of the wave

<h3>What is angular frequency of a wave?</h3>

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue colored graph; at y = 1, x = -2 cm

1 = cos(2 - C)

(2 - C) = cos^(1)

(2 - C) = 0

C = 2

Thus, the value of parameter C for the function in the figure is 2.

Learn more about phase angle here: brainly.com/question/16222725

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Experiment: Limestone
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When these bonds are destroyed, a reaction occurs. ... Vinegar reacting with limestone breaks the bonds of calcium carbonate and acetic acid.
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3 years ago
There is little vertical air movement in the __ because of the cold, heavy air at the bottom of the layer.
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The correct answer is the Mesosphere. <span>Mesosphere </span><span>is the layer of the Earth's atmosphere that is directly above the stratopause and directly below the mesopause.</span>
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Materials that are soft and porous will absorb energy causing a decrease in amplitude and energy of the sound. This is called __
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Damping of the sound
5 0
3 years ago
An object is 15 cm in front of a diverging lens with a
Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

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

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0
3 years ago
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