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klio [65]
3 years ago
10

A child holds one end of a 33.0-meter long rope in her hand and moves it up-and-down to produce a sinusoidal wave by moving her

hand from 8.00 cm above her shoulder to 8.00 cm below her shoulder at a frequency of 2.00 Hz and a wavelength of 75.0 cm. If the child doubles the amplitude of her hand's motion on that same rope with the same tension in the rope, then what will the wavelength of the wave now be
Physics
1 answer:
WITCHER [35]3 years ago
6 0

Answer:

Wavelength=75 cm.

The wavelength well remain unchanged which is 75 cm.

Explanation:

The formula which will help us to answer the question is:

V=f*λ

Where:

V is the velocity

f is the frequency of wave

λ is the wave length

Now:

λ=V/f    Eq (1)

The equation show's that wavelength is independent of the amplitude but it depends on the frequency and the velocity with which wave is moving.

The wavelength well remain unchanged which is 75 cm.

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In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
Is glass weigh more than wood
inn [45]
It depends how large the object is. But, if it is 2 by 4, most likely the wood sieges more than the glass. Hope this helps.
6 0
3 years ago
Read 2 more answers
ich of the following is not accurate when describing solids? A. The amount of pressure exerted by a solid is solely dependent on
Dafna11 [192]
The amount of solid does not affect how you are describing the solid so a is the answer
5 0
3 years ago
Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
Harrizon [31]

1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.

The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().

The Rydberg formula is used to determine the energy change.

Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.

aaΔE=R(1n2f−1n2i) aa

were

2.17810-18lJ is the Rydberg constant.

The initial and ultimate energy levels are ni and nf.

As a change of pace from

n=5 to n=3 gives us

ΔE

=2.178×10-18lJ (132−152)

=2.178×10-18lJ (19−125)

=2.178×10-18lJ×25 - 9/25×9

=2.178×10-18lJ×16/225

=1.549×10-19lJ

Learn more about Rydberg formula here-

brainly.com/question/13185515

#SPJ4

8 0
2 years ago
Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th
Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0
3 years ago
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