Tom has a mass of 70.0 kg and Sally has a mass of 50.0 kg. Tom and Sally are standing 20.0 meters apart on the dance floor. Sally looks up and sees Tom. She feels an attraction. If the attraction is gravitational, find its size. Assume that both Tom and Sally can be replaced with sephrical masses.
Please help me with this problem.
I'm having trouble understanding it.
2006-12-08 14:31:54 · 6 answers · asked by swimmertommy 1 in Science & Mathematics ➔ Physics
ask you friend
2006-12-08 14:35:09 · answer #1 · answered by Anonymous · 1⤊ 7⤋
It is a silly way of asking what the gravitational attraction between two spherical masses, one 70kg and the other 50kg, 20 meters apart is.
Equation for gravitational attraction between two spherical masses is:
F=G*m1*m2/d^2
where
G = 6.67*10^-11 = universal gravitational constant
m1 = mass of the first body
m2 = mass of the second body
d = the distance between the two bodies
Hope that helps.
m
2006-12-08 14:44:41 · answer #2 · answered by Mukluk 2 · 3⤊ 0⤋
5.84e-10 N
It is asking for the gravitational pull between two people, which is naturally going to be very small.
The equation goes like this:
F=Gm(1)m(2)/r^2
G is the gravitational constant, 6.67x10^-11 m^3 / kg s^2
m(1) and m(2) are the masses of the two objects.
r is the distance between the two objects.
the equation would look like this: F=(6.67e-11 x 70 x 50)/20^2
the force is F, so it is equal to 5.84 x 10^-10
2006-12-08 14:37:51 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Tom Sally
2017-02-24 03:11:45 · answer #4 · answered by tekchand 3 · 0⤊ 0⤋
F = GMm/(r)^2
G = 6.67*10^(-11)
M = 70
m = 50
r = 20
F = 6.67*10^(-11) * 70 * 50/(20)^2
F = 6.67*10^(-11)*3500/400
F = 6.67*10^(-11)*35/4
F = 233.45/400000000000
F = 0.000000000583625 N
2006-12-08 14:56:39 · answer #5 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
Fgravitational= 6.67x10^-11m1m2 / d^2
Fgravitational= 6.67x10^-11(70.0kg)(50.0kg) / (20m)^2
Fgravitational= 5.83x10^-10
2006-12-08 14:42:31 · answer #6 · answered by 7 · 1⤊ 0⤋