Answer:
Height, h = 16.67 m
Explanation:
We have,
Mass of a squirrel is 0.765 kg.
He jumps off the tree and hits the ground with 125 joules of energy.
It is required to find the height up on the tree the squirrel was when it jumped.
The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

h is height up on the tree the squirrel was when it jumped

So, the squirrel will go to a height of 16.67 m.
<h2>
Answer: 7020.117 m/s</h2>
Explanation:
The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting, in this case, the Earth
.
the radius of the orbit (measured from the center of the planet to the satellite).
This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth
and the altitude of the satellite above the Earth's surface
.
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).
Now, rewriting equation (1) with the known values:
Answer:
- 210 rad/s²
Explanation:
n = frequency of rotation = 3400/60 = 170/3 per sec.
angular velocity ω ( 0 ) at time 0 = 2π n = 2π x 170/3
angular velocity at time t = ω(t) = 0
now, ω²( t) = w²(o) + 2α Φ ( α = angular acceleration and Φ = angular displacement) = 2π x 48 rad.
0 = ( 2π x 170/3 )² + 2α x 48 x 2π
α = - (2π x 170 x 170 )/ (3 x 3 x 2 x 48 ) = 210 rad / s²
Answer:
Area of the plates of a capacitor, A = 0.208 m²
Explanation:
It is given that,
Charge on the parallel plate capacitor, 
Electric field, E = 3.1 kV/mm = 3100000 V/m
The electric field of a parallel plates capacitor is given by :



A = 0.208 m²
So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.