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Nana76 [90]
3 years ago
4

In which type of wave do the particles in a medium move parallel to the direction that the wave moves?

Physics
2 answers:
leva [86]3 years ago
4 0
The type of wave that the particles in a medium move parallel to the direction that the wave moves is longitudinal waves. The correct answer is D. 
Ierofanga [76]3 years ago
4 0

Answer: The correct answer is Option D.

Explanation:

Ocean waves are the waves that are causes by the crashing of water on the beach.

Surface waves are defined as the seismic waves that travel across the surface of the Earth.

Transverse waves are the waves in which the particles of the medium travel perpendicularly to the direction of the wave.

Longitudinal waves are the waves in which the particles of the medium travel in the direction of the wave.

Hence, the correct answer is Option D.

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A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will
gtnhenbr [62]
1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
x(t)=v_0t
y(t)=h- \frac{1}{2}gt^2
where the motion on the x-axis is a uniform motion with constant speed v_0=18.1 m/s, while the motion on the y-axis is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2, and initial height h=1.85 m.

First of all, we can find the time t at which the ball reaches the ground by requiring y(t)=0:
0=h- \frac{1}{2}gt^2
t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }=  0.61 s

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one: 
v_x=18.1 m/s
On the y-axis, the velocity is given by
v_y(t)=gt
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s
And the speed of the ball is the magnitude of the resultant of the two components:
v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s
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