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3 years ago

How is wave movement affected by the various mediums it may travel through?

Physics

1 answer:

Juliette [100K]3 years ago

5 0

Answer:

speed is affected by the medium in wich the wave travels.

Explanation:

The speed of a wave is affected when traveling in different medium, depending on the medium in which the wave travels it can go faster or slower, or not even be able to travel in that medium.

For light waves, the denser the medium through which the wave travels, the slower it will go through it, as there is more resistance to the light particles.

And in waves like sound, which instead of particles is a vibration or disturbance in the medium, they travel faster in denser mediums such as metal, because the disturbance is transmitted more efficiently thanks to how close together that the particles are in the metal, This is why sound does not travel in a vacuum; there are not even particles to transmit the vibration.

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Please help me, I will give BRAINLIEST to the best answer

sdas [7]

Explanation:

a. A grocery bag as you lift it up

b. A crane moving dirt......

d. A crate as you push it along the floor

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3 years ago

Read 2 more answers

a golfer tees off and hits a golf ball at a speed of 31 m/s and an angle of 35 degrees. how far did the ball travel before hitti

poizon [28]

Ans: R = Ball Travelled = 92.15 meters.

Explanation:
First we need to derive that formula for the "range" in order to know how far the ball traveled before hitting the ground.

Along x-axis, equation would be:
$x = x_o + v_o_xt + \frac{at^2}{2}$

Since there is no acceleration along x-direction; therefore,
$x = x_o + v_o_xt$

Since $v_o_x = v_ocos \alpha$ and $x_o$ =0; therefore above equation becomes,

$x = v_ocos \alpha t$ --- (A)

Now we need to find "t", and the time is not given. In order to do so, we shall use the y-direction motion equation. Before hitting the ground y ≈ 0 and a = -g; therefore,

=> $y = y_o + v_o_yt - \frac{gt^2}{2}$
=> $t = \frac{2v_o_y}{g}$

Since $v_o_y = sin \alpha$ ; therefore above equation becomes,
$t = \frac{2v_osin \alpha }{g}$

Put the value of t in equation (A):

(A) => $x = v_ocos \alpha \frac{2v_osin \alpha }{g}$

Where x = Range = R, and $2sin \alpha cos \alpha = sin(2 \alpha )$ ; therefore above equation becomes:

=> $R = (v_o)^2 *\frac{sin(2 \alpha )}{g}$

Now, as:
$v_o = 31 m/s$

and $\alpha = 35$ °
and g = 9.8 m/(s^2)

Hence,
$R = (31)^2 *\frac{sin(2 *35 )}{9.8}$

Ans: R = 92.15 meters.

-i

7 0

3 years ago

What characteristic of a planet determines the acceleration due to gravity on it?

Korvikt [17]

The mass of a planet determines the acceleration due to gravity on it. This is according to Newton's Law of Gravitation, which basically states that the more mass a body has, the greater the force of attraction it exerts on other bodies with mass near it.

The gravitational force is:

F = GMm/r², where G is a constant, r is the distance between large mass M and small mass m.

Considering the fact that acceleration is force per unit mass, if we divide gravitational force by the small mass (to get force per unit mass), we see the dependence mathematically:

a = GM/r²

7 0

3 years ago

skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If

scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

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3 years ago

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

Elan Coil [88]

Make the base of the building zero. Then the initial distance is 100m, final distance unknown x. Use gravity, time and initial velocity to solve for final distance.
x - 100 = (0)(5) +(1/2)(-9.81)(5^2)
x - 100 = 0 - 122.625
x = -122.625 + 100
x = -22.625 m below ground

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3 years ago