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erastovalidia [21]
2 years ago
5

A rope is shaken and produces 2 waves each second. Calculate the time period of the rope waves.

Physics
1 answer:
motikmotik2 years ago
7 0

Answer:

0.5 s

Explanation:

From the question given above, the following data were obtained:

Number of circle (n) = 2

Time (t) = 1 s

Period =?

Period of a wave is simply defined as the time taken to make one complete oscillation. Mathematically, it can be expressed as:

T = t / n

Whereb

T => is the period

t => is the space time

n => is the number of circle or oscillation.

With the above formula, we can obtain the period of the wave as follow:

Number of circle (n) = 2

Time (t) = 1 s

Period =?

T = t / n

T = 1 / 2

T = 0.5 s

Thus, the period of the wave is 0.5 s

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Force is (mass × acceleration) measured in Newton

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If the absolute pressure of a gas is 550.280 kPa, its gage pressure is A. 101.325 kPa. B. 651.605 kPa. C. 448.955 kPa. D. 277.28
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Answer:

Option C is the correct answer.

Explanation:

Absolute  pressure is sum of gauge pressure and atmospheric pressure.

That is

               P_{abs}=P_{gauge}+P_{atm}

We have

          P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa

Substituting

         P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa

Option C is the correct answer.

7 0
3 years ago
Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen
muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass m=0.20kg

Velocity v=4m/s

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

 M_{a1}=mV

 M_{a1}=0.2*4

 M_{a1}=0.8

Final Momentum

 M_{a2}=-0.8kgm/s

Therefore

 \triangle M_a=-1.6kgm/s

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

 M_{b1}=mV

 M_{b1}=0.2*4

 M_{b1}=0.8

Final Momentum

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Therefore

 |\triangle M_a|>|\triangle Mb|

Option A

4 0
3 years ago
What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
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Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

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The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

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7 0
2 years ago
The first law of motion applies<br> to what?
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First law of motion<span>- sometimes referred to as the </span>law<span> of inertia. An object at rest stays at rest and an object in </span>motion<span> stays in </span>motion<span> with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
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