Answer:
F(Mars) = 2 G m M / (4 R)^2 force of Sun on Mars
F(Merc) = G m M / R^2 force of force of Sun on Mercury
R = distance of Sun from Mercury, m = mass of Mercury
F(Merc) / F(Mars) = 4^2 / 2 = 8
Answer:
So the force of attraction between the two objects is 3.3365*10^-6
Explanation:
m1=10kg
m2=50kg
d=10cm=0.1m
G=6.673*10^-11Nm^2kg^2
We have to find the force of attraction between them
F=Gm1m2/d^2
F=6.673*10^-11*10*50/0.1^2
F=3.3365*10^-8/0.01
F=3.3365*10^-6
The relationship between the distance covered, initial and final speeds, and time can be expressed through the equation,
First equation,
2ad = Vf² - Vi²
Substituting the known values,
2(a)(0.230 km) = (70 km/h)² - (40 km/h)²
The value of a from the equation is 7173.92 km/h².
Second equation,
d = (Vi)(t) + 0.5at²
Substituting the known values,
0.230 km = (40 km/h)(t) + (0.5)(7173.92 km/h²)(t²)
The value of t from the equation is 4.1818 x 10^-3 hours which is also equal to 0.2509 minutes or 15 seconds.
Answer: 15 seconds