It would be the electromagnet because that is how it spins
Answer:
891 excess electrons must be present on each sphere
Explanation:
One Charge = q1 = q
Force = F = 4.57*10^-21 N
Other charge = q2 =q
Distance = r = 20 cm = 0.2 m
permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)
Using Coulomb's law,
F=[1/4pieo]q1q2/r^2
F = [1/4pieo]q^2 / r^2
q^2 =F [4pieo]r^2
q = r*sq rt F[4pieo]
q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]
q = 1.42614*10^ -16 C
number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19
n =891
891 excess electrons must be present on each sphere
Answer:
maybe try searching it up
<span>Answer: Va = 7,625 m/s
Vb = 7,404 m/s
Given:
A = 486,000 m
B = 901,000 m
G = 6.67428E-11 m^3/kg-s^2
M = 5.9736E+24 kg
r = 6,371,000 m
Recall that you need the actual orbital distance from the *center* of the Earth, giving radius plus altitude:
rA = 6,857,000 m
rB = 7,272,000 m
Equation:
V = SQRT { GM / r }
Solve for A
Va = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (6,857,000 m) }
Va = SQRT { [ 3.9869 m^3/s^2 ] / (6,857,000 m) }
Va = SQRT { 58,144,202 m^2/s^2 }
Va = 7,625 m/s
Solve for B
Vb = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (7,272,000 m) }
Vb = SQRT { [ 3.9869 m^3/s^2 ] / (7,272,000 m) }
Vb = SQRT { 54,826,016 m^2/s^2 }
Vb = 7,404 m/s</span>
