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ratelena [41]
3 years ago
15

You are creating waves in a rope by shaking your hand back and forth. Without changing the distance your hand moves, you begin t

o shake it faster and faster. What happens to the amplitude, frequency, period, and velocity of the wave?
Physics
1 answer:
scoundrel [369]3 years ago
7 0

Answer:

<u>Amplitude - remains the same</u>

<u>Frequency - increases</u>

<u>Period - decreases</u>

<u>Velocity - remains the same.</u>

<u />

Explanation:

The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.

The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.

The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.

The velocity of a wave depends on the medium in which it is travelling. Your wave was previously travelling in air and the new wave is also travelling in the same medium so the velocity of the wave remains unchanged.

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Is it True or false a muscle strain occurs when a ligament is torn away from the bone
Anit [1.1K]
I think the answer its false i dont know how to explain that but it is false that what my teacher told me 
6 0
3 years ago
An object is 10 cm from the mirror, its height is 1 cm and the focal length is 5 cm. What is the image height? (Indicate the obj
boyakko [2]

The image height is -10 cm (the image is upside down)

Explanation:

We can solve the problem by using the mirror equation:

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

where:

f = 5 cm is the focal length of the mirror

p = 10 cm is the distance of the object from the mirror

q is the distance of the image from the mirror

Solving for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\\\rightarrow q= 10 cm

So, the distance of the image from the mirror is 10 cm.

Now we can find the image height by using the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the height of the image

y = 1 cm is the height of the object

and using

p = 10 cm

q = 10 cm

We find the size of the image:

y' = -\frac{qy}{p}=-\frac{(10)(1)}{10}=-10 cm

where the negative sign indicates that the image is upside down.

#LearnwithBrainly

8 0
3 years ago
Find the wavelength of the third line in the lyman series, and identify the type of em radiation.
Natalija [7]

The wavelength of the third line in the Lyman series, and identify the type of EM radiation

In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.

1 / lambda = R(h)* ( \frac{1}{(n1)^{2} } -   \frac{1}{(n2)^{2} })

                 = 109678 ( \frac{1}{1^{2} } -  \frac{1}{3^{2} } )

                 = 109678 (8/9)

   Lambda = 9 / (109678 * 8 )

                  = 102.6 * 10^{-9} m = 102.6 nm

To learn more about Lyman series here

brainly.com/question/5762197

#SPJ4

8 0
2 years ago
23 is 2% of what<br> number?
8_murik_8 [283]

Answer:

1;150

Explanation:

2% × ? = 23

? =

23 ÷ 2% =

23 ÷ (2 ÷ 100) =

(100 × 23) ÷ 2 =

2,300 ÷ 2 =

1,150;

7 0
3 years ago
Read 2 more answers
How do I solve this​
Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
4 years ago
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