Answer:D increase in frequency
Explanation:
Explanation:
The electric field at a distance r from the charged particle is given by :
k is electrostatic constant
if r = 2 m, electric field is given by :
If r = 1 m, electric field is given by :
Dividing equation (1) and (2) we get :
So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.
Bc the equator is constantly facing the sun while the earth rotates
Hi,
Recall the formula V=d/t, where V stands for velocity or speed, d stands for distance and t stands for time. By substituting your values you get: V=50.0km/2.5h, which equals to 20km/h.
'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.
For the stone dropped from the 50m tower:
H = +50 - (1/2) G T²
For the stone tossed upward from the ground:
H = +20T - (1/2) G T²
When the stones' paths cross, their <em>H</em>eights are equal.
50 - (1/2) G T² = 20T - (1/2) G T²
Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:
50 = 20T Isn't that incredible ? ! ?
Divide each side by 20 :
<u>2.5 = T</u>
The stones meet in the air 2.5 seconds after the drop/toss.
I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?
Their height is +50 - (1/2) G T² = 19.375 meters
The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.
