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3 years ago

What will occur when the waves peak at the same place at the same time?

1 answer:

5 0

Hello!

This is a matter of superposition.
When the waves peak at the same time and place, they produce constructive interference, meaning the waves interact together in a positive way, to make a wave with Amplitude of both waves added together. When the peaks differ however, at the same time and place, then it is destructive interference and the waves essentially cancel each other out.

Hope this helps. Any questions please just ask. Thank you kindly.

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newton’s second law states that f=m x a (force is mass times acceleration). which example would have the greatest acceleration

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Answer:

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3 years ago

What is transmitted by all waves? energy mass matter sound

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Answer: energy

Explanation: A wave is a disturbance that transfers energy as it travels through medium.

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3 years ago

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A planet has a period of revolution about the sun equal to T and a mean distance from the sun equal to R. T2 varies directly as

Tamiku [17]

Answer:

T² ∝ R³

Explanation:

Given data,

The period of revolution of the planet around the sun, T

The mean distance of the planet from the sun, R

According to the III law of Kepler, " Law of Periods' states that the square of the orbital period to go around the sun once is directly proportional to the cube of the mean distance between the sun and the planet.

T² ∝ R³

$\frac{T^{2}}{R^{3}} = Constant$

From the above equation it is clear that T² varies directly as the R³.

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3 years ago

A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise.

Lera25 [3.4K]

The true statement about the CD is:

<h3><em>b. No net torque acts on it at all.</em></h3>

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<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

<em>a = Centripetal Acceleration ( m/s² )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

$\texttt{ }$

Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

<em>F = Centripetal Force ( m/s² )</em>

<em>m = mass of Particle ( kg )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

Let us now tackle the problem !

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<em>Complete Question:</em>

<em>A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise. Which of the following statements about the CD is true?</em>

<em>a. A net torque acts on it clockwise to keep it moving</em>

<em>b. No net torque acts on it at all.</em>

<em>c. A net torque acts on it counterclockwise to keep it moving</em>

<u>Given:</u>

angular velocity = ω = 5.0 revolutions per second

<u>Asked:</u>

net torque = Στ = ?

<u>Solution:</u>

Constant angular velocity → angular acceleration = α = 0 rad/s²

$\Sigma \tau = I \alpha$

$\Sigma \tau = I (0)$

$\Sigma \tau = 0 \texttt{ Nm}$

$\texttt{ }$

<h3>Conclusion:</h3>

The true statement about the CD is:

<em>b. No net torque acts on it at all.</em>

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

#LearnWithBrainly

8 0

3 years ago

A 26.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal. If

Nuetrik [128]

Answer:

Force exerted by the hinge on the beam = 109.24N

Explanation:

Weight = mg = 26 x 9.81 = 255.06 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, there are 3 vertical forces acting on the beam. These are;

- The downward force which is the weight of the beam.

- The vertical components of the tension in the cable.

-The force that hinge exerts on the beam are the upward forces.

Hence, for the beam to remain horizontal, the sum of the upward forces must be equal to the weight of the beam.

For us to determine the vertical component of the tension in the cable, we will do a torque problem. Let the pivot point be at the hinge. Let’s assume that the length of the beam is L. The vertical component of the tension in the cable will produce clockwise torque while the weight of the beam will produce counter clockwise torque.

Tbus;

Clockwise torque = TL sin 61

Since the center of mass of beam is at the middle of the beam, the distance from the hinge to the weight of the beam is L/2.

Counter clockwise torque = WL/2

Thus;

TL sin 61 = WL/2

L will cancel out.

T sin 61 = 255.06/2

T x 0.8746 = 127.53

T = 127.53/0.8746 = 145.82 N

Now, the equation to determine the vertical component of the force that the hinge exerts on the beam is given as;

T + F = W

Thus;

145.82 + F = 255.06

F = 255.06 - 145.82 = 109.24 N

8 0

3 years ago