Parta a.
Equation: F = G*m1*m2/d^2
Where
F = 32 N
G = 6.67*10^-11 N.m^2/kg^2
m1 = 9.0*10^13kg
m2 =370 kg
d = distance that separate the center of the two objects.
d^2 = G*m1*m2 / F = 6.67*10^-11 N.m^2/kg^2 * 9.0*10^13 kg *370 kg / 32N = 69,409.69 m^2
d = √69,409.69m^2 = 263.5 m
Part B.
The gravitational field of the comet is g = G*m1/d^2
Notice that it does not depend on the mass of other objects.
Notice also that I will use a distance of 5.0 * 10^3 km, because I think that that is the number that you intended to write in the part b. If that is not the number you can put the right number instead because the solution is written step by step.
g = (6.67*10^-11 N*m^2/kg^2)*(9.0*10^13kg)/(5.0*10^3*10^3m)^2 = 2.4*10^-4 N/kg = 2.4*10^-4 m/s^2
Answer:
The distance is 54.6 m
Explanation:
Given that,
Mass = 2.0 kg
Frictional coefficient = 0.21
Initial velocity = 15 m/s
We need to calculate the acceleration
Using formula of frictional force
We need to calculate the acceleration
We need to calculate the initial velocity
Using equation of motion
Put the value
Hence, The distance will be 54.6 m.
Answer:
The geosphere consists of the solid Earth and the atmosphere consists of the gaseous components in the air. Thus, the answer is C.
Explanation:
180 = 2x+x+21
159 = 3x
x = 53
Angle A = 53°
Angle B = 53°
Angle C = 53+21 = 74°
d = distance between the two point charges = 60 cm = 0.60 m
r = distance of the location of point "a" where the electric field is zero from charge 
between the two charges.
= magnitude of charge on one charge
= magnitude of charge on other charge
= 3
= Electric field by charge at point "a"
= Electric field by charge at point "a"
Electric field by charge at point "a" is given as
= k /r²
Electric field by charge at point "a" is given as
= k /(d-r)²
For the electric field to be zero at point "a"
=
k /(d-r)² = k /r²
/(d-r)² = 3 /r²
1/(0.60 - r)² = 3 /r²
r = 0.38 m
r = 38 cm
