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Eduardwww [97]
3 years ago
15

the time required for one cycle, a complete motion that returns to its starting point, it called the_____ medium frequency perio

d periodic motion
Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

The correct answer to the following question will be "Period".

Explanation:

The Period seems to be the time deemed necessary for such a perfect cycle of vibration to transfer a particular moment. Because as the amplitude of the wave raises, the wavelength falls.

It is denoted by "T" and its formula will be:

⇒  T  = \frac{1}{F}

Where, T = Period

            F = Frequency

The other given choices are not related to the given circumstances. So that the above would be the right answer.

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PLS HELP! A boy drops a ball from an observation tower. The ball hits the ground in 3.0 s. What is the ball's velocity at the ti
Levart [38]

Answer:

v = u + at

v = 0 + 9.8 × 3

v = 29.4 m/sec

4 0
3 years ago
A train with proper length L has clocks at the front and back. A photon is fired from the front to the back. Working in the trai
tamaranim1 [39]

Explanation:

In train's rest frame, the speed of photon is c and the proper length of the train is L. The time taken by the photon to cross the train is t=\frac{L}{c}

In ground frame, the speed of the photon is given as follows:

v_{x}=\frac{v_{x}+v}{1+\frac{v_{x} \cdot v}{c^{2}}}

=\frac{c+v}{1+\frac{c v}{c^{2}}} \\=c

The speed of light or photon remains same in every frame of reference.

Now, the speed of train is very less as compared to the speed of photon so that v So that, \frac{v}{c} \ll 1

The length contraction in the ground frame is given as follows:

L^{\prime}=L \sqrt{1-\frac{v^{2}}{c^{2}}}

=L

Time taken by the photon to travel the length of the train in ground frame is .

7 0
3 years ago
A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from
Alik [6]

Answer:

Speed is same as that before it entered glass.

Explanation:

Given:

A light ray enters and passes through the glass as shown in the diagram.

We have to analyze its speed.

Speed of light in air is 3\times 10^8\ ms^-^1 and speed of light in glass is 2.25\times 10^8\ ms^-^1

Whenever a light ray enters a glass block or slab there is bending of light at the interface of the two media.

So speed of light will decrease in glass medium and again it passes to the air.

So

Speed of light in air will again increase or will be equivalent to the earlier speed when it was entering the glass block.

Finally

Speed is same as that before it entered glass as it in the same medium (air).

6 0
3 years ago
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

5 0
3 years ago
1. Give three examples, from the lab, where potential energy was converted to kinetic energy: ​
Lubov Fominskaja [6]

Answer:

A book on a table before it falls.

A yoyo before it is released.

A raised weight.

Explanation:

These are all examples of potential energy. So I hope you can find something that is comparable from the lab.

3 0
3 years ago
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