Answer:
Explanation:
The apparent brightness follows an inverse square law, therefore we can write:
where I is the apparent brightness and r is the distance from the Sun.
We can also rewrite the law as
(1)
where in this problem, we have:
apparent brightness at a distance 
, where
million km
We want to estimate the apparent brightness at 
, where is ten times , so
Re-arranging eq.(1), we find 
:
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Answer:
3.88 * 10^(-15) J
Explanation:
We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.
First, we get the potential and potential energy:
Electric potential = E * r
E = electric field
r = distance between plates
Potential = 2.2 * 10^6 * 0.011
= 2.42 * 10^4 V
The relationship between electric potential and potential energy is:
P. E. = q*V
q = charge of electron = 1.602 * 10^(-19) C
P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)
P. E. = 3.88 * 10^(-15) J
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
