To answer this question, we should know the formula for the terminal velocity. The formula is written below:
v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient
Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²
v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
I don't think so. How are you supposed to write the answer?
Answer:
5.7 x 10^12 C
Explanation:
Let the charge on earth and moon is q.
mass of earth, Me = 5.972 x 10^24 kg
mass of moon, Mm = 7.35 x 10^22 kg
Let d be the distance between earth and moon.
the gravitational force between them is
The electrostatic force between them is
According to the question
1 % of Fg = Fe
q = 5.7 x 10^12 C
Thus, the charge on earth and the moon is 5.7 x 10^12 C.
Average speed = total distance travelled ÷ total time taken
AS = (75km + 68km) ÷ (1hr + 2hr)
As = 143km ÷ 3hr
AS = 47.66667 km/hr
AS = 47.7 km/hr (3sf)
