(a) 0.448
The gravitational potential energy of a satellite in orbit is given by:

where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):
r = R + h
We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

and so, substituting:

We find

(b) 0.448
The kinetic energy of a satellite in orbit around the Earth is given by

So, the ratio between the two kinetic energies is

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.
(c) B
The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

For satellite A, we have

For satellite B, we have

So, satellite B has the greater total energy (since the energy is negative).
(d) 
The difference between the energy of the two satellites is:

The change in momentum of an object equals the impulse applied to it
Answer:
v= s/t
Explanation:
250 km/ h =69.44m/s
S1=2 times 69.44 ≈ 139m
Next 2.5 seconds:
S2 = 100m
Average speed:
v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h
Answer:
1.67 A
Explanation:
Given that,
→ Power (P) = 400 W
→ Potential difference (V) = 240 V
→ Current (I) = ?
The amount of current drawn will be,
→ P = V × I
→ I = P/V
→ I = 400/240
→ I = 1.66666666667
→ [ I = 1.67 A ]
Hence, the current drawn 1.67 A.
Answer:
640 nanometer setara dengan 6.4e-7 meter