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Vitek1552 [10]
3 years ago
13

If a ball is kicked horizontally off of a building at 8.0 m/s, what is the horizontal velocity (vx) after 3

Physics
1 answer:
nadya68 [22]3 years ago
6 0

The characteristics of the projectile launch allows to find the result for the velocity on the x axis is:

  • In the horizontal axis there is no acceleration and therefore the speed must be kept constant.
  • The horizontal speed is  vₓ = 8.0 m / s

In projectile launching we have an application of kinematics to motion in two dimensions near the surface of the Earth.

In this case, on the vertical axis we have an acceleration directed towards the center of the planet and called the gravity acceleration.

In the horizontal axis or x axis there is not acceleration, except in special cases that we tone is how much the resistance of the air.

Consequently in the horizontal axis there is no acceleration and therefore the speed must be kept constant, the result for the horizontal speed is:

           vₓ = 8.0 m / s

Learn more here: brainly.com/question/17228361

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kondaur [170]

<u>Absorb:</u>
Shine a light with known intensity into the object. 
Measure the intensity of the light that comes out.
Any light that goes in but doesn't come out the other side
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(Note:  You also need to be aware of any light that bounces back from
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4 0
3 years ago
The electrons in a particle beam each have a kinetic energy K. What is the magnitude of the electric field that will stop these
harina [27]

Answer:

Electric field magnitude

E = K/qd

Where

K = kinetic energy of electron

d = electron distance

q = charge

Explanation:

Given the relationship between workdone and energy

Work-energy theorem:

Net workdone = Energy change

W = ∆E

In this case

W = ∆K.E

And,

∆K.E = K(final) - K(initial)

To stop the kinetic energy | K(final) = 0

K(initial) = K (given)

∆K.E = 0 - K = -K

Let the electric force on the electron has magnitude F.

And

W = -Fd = ∆K.E = -K

-Fd = -K

F = K/d .....1

The magnitude of the electric field E that can stop these electron in a distance d:

E = F/q ......2

Where q is the charge on electron.

substituting equation 1 to 2

E = (K/d)/q = K/qd

E = K/qd

3 0
4 years ago
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Answer: the waves probably hits an open cavity

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