Answer:11 km/s
Explanation:
Given
Escape velocity at the surface of earth is 11 km/s
Escape velocity is given by
Escape velocity at the surface of earth
--------------------1
If Escape velocity is three times and the radius is also the three times
i.e. 
As long as it sits on the shelf, its potential energy
relative to the floor is . . .
Potential energy = (mass) x (gravity) x (height) =
(3 kg) x (9.8 m/s²) x (0.8m) = <u>23.52 joules</u> .
If it falls from the shelf and lands on the floor, then it has exactly that
same amount of energy when it hits the floor, only now the 23.52 joules
has changed to kinetic energy.
Kinetic energy = (1/2) x (mass) x (speed)²
23.52 joules = (1/2) x (3 kg) x (speed)²
Divide each side by 1.5 kg : 23.52 m²/s² = speed²
Take the square root of each side: speed = √(23.52 m²/s²) = <em>4.85 m/s </em> (rounded)
Earths Rotation
Earth's rotation is the rotation of Planet Earth around its own axis. Earth rotates eastward, in prograde motion. As viewed from the north<span> pole star Polaris, Earth turns counter</span>clockwise<span>. </span>
"AB84"
-1- was created in the 1600 by william gilbert
-2-When the charge is positive, electrons in the metal of the electroscope are attracted to the charge and move upward out of the leaves. This results in the leaves to have a temporary positive charge and because like charges repel, the leaves separate. When the charge is removed, the electrons return to their original positions and the leaves relax
3-
An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x 
m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x 
s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x x 3 x )/2
=120m
