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Elis [28]
3 years ago
11

 

Physics
2 answers:
Shtirlitz [24]3 years ago
7 0

Answer:

Your correct answer is W

Explanation:

Hope this helps :) -Mark Brainiest Please :)

Leto [7]3 years ago
6 0

Answer:

w

Explanation:

..

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1. Any push or pull is called
timama [110]

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b  ok

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Compared to yellow light,orange light has
ArbitrLikvidat [17]
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4 years ago
How does wavelength change depending on whether the pipe is open or closed?
il63 [147K]

Answer:

* in a tube with both ends open

        λ = 2L / n                    n = 1, 2, 3,…

* tube with one end open and the other closed.  

         λ = 4L / n                     n = (2n ’+ 1 )

Explanation:

When a sound wave enters a pipe, a resonance process occurs, whereby only some wavelengths can occur.

 

* in a tube with both ends open

in this case there are maximums at each end, so if the length of the had is L

              λ = 2L 1st harmonic

              λ = 2 L / 2 2nd harmonic

              λ = 2L / 3

              λ = 2L / n                    n = 1, 2, 3,…

* In the case of a tube with one end open and the other closed.

At the open end there is a belly and at the closed end a node

              λ = 4L 1st harmonic

              λ = 4L / 3 3rd harmonic

              λ = 4L / 5

              λ = 4L / n                  n = 1, 3, 5,… odd

                                       n = (2n ’+ 1 )      where n’ are all integers

4 0
3 years ago
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
8 0
3 years ago
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