We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1) Radius = √6.674e-11*5.972e24/8 = 7058 kms Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2 Radius=√6.674e-11*5.972e24/4.9 = 9019 kms Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
Explanation:
6000 years = 6000 x 365 x 24 x 60 x 60
= 1.892 x 10¹¹ second
gain is 1 second
1 second is equivalent to 9.193 × 10⁹ oscillations .
In 1.892 x 10¹¹ second , change in oscillation is 9.193 × 10⁹ oscillation
in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹
= 4.859 x 10⁻² oscillations .
<span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:
R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km
here, the speed of the satellite is:
v = sqrt(R*9.00m/s^2) = 7730 m/s
the time it would take the satellite to complete one full rotation is:
T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h
Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)</span>
Answer:
b) q large and m small
Explanation:
q is large and m is small
We'll express it as :
q > m
As we know the formula:
F = Eq
And we also know that :
F = Bqv
F = 
Bqv =
or Eq =
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.
E = mc^2
m = e/c^2
m = 2.7*10^16/(300000^2)
m = 300000
