(A) 14.8 g → 1.48 × 10^7 µg
The prefix µ means “× 10^(-6)”, so 1 µg = 1 × 10^(-6) g
The conversion factor (CF) = 1 µ g/1 × 10^(-6) g
∴ Mass = 14.8 g × [1 µg/1 × 10^(-6) g] = 1.48 × 10^7 µg
(B) 3.72 g → 3.72× 10^(-3) kg
The prefix k means “× 10^3”, so 1 kg = 1 × 10^3 g
CF = 1 kg/1 × 10^3 g
∴ Mass = 3.72 g × [1 kg/1 × 10^3 g] = 3.72× 10^(-3) kg
(C) 7.5 × 10^4 J → 75 kJ
CF = 1 kJ/1 × 10^3 J
∴ Energy = 7.5 × 10^4 J × [1 kJ/1 × 10^3 J] = 3.72× 10^(-3) kJ = 7.5 × 10^1 kJ
= 75 kJ
Refer to the diagram shown below.
In geosynchronous orbit, the satellite makes one circular revolution at the equator in 1 day.
From tables, obtain
r = 42,164 km = 4.2164 x 10⁷ m, the earth's radius
g = 9.8 m/s², acceleration due to gravity.
Note that 1 day = 86400 s
The angular velocity of rotation is
ω = (2π)/86400 = 7.2722 x 10⁻⁵ rad/s
The centripetal acceleration on the satellite is
a = rω²
= (4.2164 x 10⁷ m) * (7.2722 X 10⁻⁵ rad/s)²
= 0.223 m/s²
The weight of the 2000 kg satellite is
W = m(g - a)
= (2000 kg) * (9.8-0.223 m/s²)
= 19154 N
= 19.154 kN
Answer: 19 kN (nearest integer)