Answer:
t = 5.56 ms
Explanation:
Given:-
- The current carried in, Iin = 1.000002 C
- The current carried out, Iout = 1.00000 C
- The radius of sphere, r = 10 cm
Find:-
How long would it take for the sphere to increase in potential by 1000 V?
Solution:-
- The net charge held by the isolated conducting sphere after (t) seconds would be:
qnet = (Iin - Iout)*t
qnet = t*(1.000002 - 1.00000) = 0.000002*t
- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:
V = k*qnet / r
Where, k = 8.99*10^9 ..... Coulomb's constant
qnet = V*r / k
t = 1000*0.1 / (8.99*10^9 * 0.000002)
t = 5.56 ms
Answer:
v = 4.10 10⁻³ m / s
Explanation:
For this exercise we will use Newton's second law where the force is magnetic
F -W = m a
As the field is directed to the north and the proton to the east, using the rule of the right hand the force is vertical upwards, the force balances the weight the acceleration is zero
F = W
q v B = m g
Let's calculate the speed
v = m g / (q B)
v = 1,673 10⁻²⁷ 9.8 / (1.6 10⁻¹⁹ 2.5 10⁻⁵)
v = 4.10 10⁻³ m / s
Gravitational forces are stronger over shorter distances, and
weaker over longer distances. That's a big part of the reason
why our bodies are attracted to the Earth with more force than
we're attracted to Jupiter, for example.
The force doesn't just get weaker in proportion to the distance.
It gets weaker in proportion to the SQUARE of the distance.
Answer:
This should be tagged for biology not physics.
Explanation:
:)
The sun exerts the strongest g-force, holding us (earth)in it's orbit, followed by the moon which affects the tides on earth.
The sun exhibits multiple temperatures from the core to the corona, with a range of over 25 million degrees Fahrenheit.