Answer:
Explanation:
Total energy of a satellite in an orbit , h height away
= - GMm /2 ( R + h )
When h = 380 km
Total energy of a satellite = 
= - 13.25 x 10¹⁰ J
When h = 190 km
Total energy of a satellite =

= - 13.63 x 10¹⁰ J
Diff
= 38 x 10⁸ J Energy will be required.
Answer:
35°C
Explanation:
q = mCΔT
2130 J = (0.200 kg) (710 J/kg/°C) (T − 20.0°C)
T = 35°C
Answer:
a)
b)
Explanation:
Given that
v(t) = 5 t i + t² j - 2 t³ k
We know that acceleration a is given as



Therefore the acceleration function a will be

The acceleration at t = 2 s
a= 5 i + 2 x 2 j - 6 x 2² k m/s²
a=5 i + 4 j -24 k m/s²
The magnitude of the acceleration will be

a= 24.83 m/s²
The direction of the acceleration a is given as

a)
b)
Answer:
μ = 0.535
Explanation:
On a level floor, normal force = weight.
N = W
Friction force = normal force × coefficient of friction.
F = Nμ
Substitute:
F = Wμ
83 = 155μ
μ = 0.535
Round as needed.
Answer:
Total energy saving will be 0.8 KWH
Explanation:
We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW
30 bulbs are of power 60 W
So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW
Total power of 80 bulbs = 1.8+5 = 6.8 KW
Total time = 3 hour
We know that energy 
Now power of each CFL bulb = 25 W
So power of 80 bulbs = 80×25 = 2000 W = 2 KW
Energy of 80 bulbs = 2×3 = 6 KWH
So total energy saving = 6.8-6 = 0.8 KWH