Answer:
The current pass the
is 
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The voltage is 
The first resistance is 
The second resistance is 
Since the resistors are connected in series their equivalent resistance is

Substituting values


Since the resistance are connected in serie the current passing through the circuit is the same current passing through
which is mathematically evaluated as

Substituting values


Answer:
1.when it is closest to the sun
2.when it is midway between its farthest
Explanation:
According to the law of Kepler's
T ² ∝ r³
T=Time period
r=semi major axis
We also know that time period T given as

v=Speed







So we can say that ,when r is more then the speed will be minimum and when r is low then speed will be maximum.
Energy cannot be created nor be destroyed
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
(1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:

The strength of the magntetic field is 4.1mT