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:
3.4 x 10⁴ m/s
Explanation:
Consider the circular motion of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of electron = 9.1 x 10⁻³¹ kg
v = radial speed
r = radius of circular path = 2 mm = 0.002 m
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
For the circular motion of electron
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Consider the linear motion of the electron :
v' = linear speed
x = horizontal distance traveled = 9 mm = 0.009 m
t = time taken =
=
= 4.5 x 10⁻⁷ sec
using the equation
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
Speed is given as
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s
Answer:
Because it is said that the earth rotates and revolves around the sun and moon so it is impossible for the earth not to be spinning.
Explanation:
Answer:
60 km/h
Explanation:
In the first part of the trip, the speed is
v = 80 km/h
while the time interval is
t = 3 h
So, the distance covered is:
d = vt = (80)(3)= 240 km
The problem states that this distance is half distance between home and the destination - so, the total distance between home and the destination is

The time taken to cover the second part of the trip is 5 h, so the total time taken is
T = 3 h + 5 h = 8 h
Therefore, the average velocity for the entire trip is
