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jolli1 [7]
3 years ago
14

A wave has a length of 8 meters and a speed of 32 meters per second. What is the frequency of the wave

Physics
1 answer:
Naddika [18.5K]3 years ago
5 0

Answer:

The frequency of wave is 4 hertz.

Explanation:

Given,

wave length = 8m

speed of sound ( v) = 32 m/s

frequency ( f) = ?

we know,

v = f × lambda

or, 32 = f × 8

or, f = 32/8

or, f = 4 Hertz.

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Answer:

Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.

Explanation:

I just learned about this 2 weeks ago actually.

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hi, Need help with triangle law of vector addition worksheet and a verifying Newtons second law worksheet?
Yuri [45]

Use Newton's second law and the free body diagram to determine the net force and acceleration of an object. In this unit, the forces acting on the object were always directed in one dimension.

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Times have changed and we are ready for situations involving two-dimensional forces. In this unit, we explore the effects of forces acting at an angle to the horizontal. This makes the force act in two dimensions, horizontal and vertical. In such situations, as always in situations involving one-dimensional network forces, Newton's second law applies.

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1 year ago
Which statement places events during the Space Race in the correct order from the earliest to the most recent?
leva [86]

Answer:

3,1,4,2

Explanation:

3 0
3 years ago
Read 2 more answers
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
Selena drew a diagram to show how current moves in a loop of wire that is placed between two magnets. At top left a piece of mag
OLEGan [10]

Answer:

B on Edge 2020

She can change the arrows so they show current traveling in opposite directions on the sides of the loop.

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Just took the test haha

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2 years ago
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