V391 Pegasi b is a star that is 4,570 light years away from Earth. What is this distance in kilometers? (One light year is about
2 answers:
You might be interested in
Answer:
1 C
Explanation:
The intensity of electric current is defined as
where
I is the current
q is the amount of charge transferred
t is the time interval during which the charge is transferred
For the lightning in this problem, we have
is the current
is the time interval
Solving the formula for q, we find the amount of charge transferred:
Answer:
v_average = 15 m / s
Explanation:
The average speed can be found in two ways,
* taking the distance traveled and divide it by the time spent
* taking the velocities in each time interval and then finding the weighted average by the time fraction
v_average = 1 / t_total ∑ vi ti
Let's apply this last equation
Total time is
t = t₁ + t₂
t = 10 + 10 = 20 min
v_average = 10/20 10 + 10/20 20
v_average = 10/2 + 20/2
v_average = 15 m / s
The average speed of the bus from Lansing to Detroit is
while the average speed of the bus from Detroit to Lansing is
The distance covered by the bus in the two trips is the same (the distance between the two cities), therefore, the average speed of the round trip can be calculated as the mean of the two speeds:
while the average speed of the bus from Detroit to Lansing is
The distance covered by the bus in the two trips is the same (the distance between the two cities), therefore, the average speed of the round trip can be calculated as the mean of the two speeds:
Answer:
Increases
Explanation:
The expression for the capacitance is as follows as;
Here, C is the capacitance, is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.
It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.
The expression for the energy stored in the parallel plate capacitor is as follows;
Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.
Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.
Therefore, the option (c) is correct.