Answer
given,
capacitance = C = 3.4-µF
inductance = L = 0.08 H
frequency is expressed as
time period
after time T/4 current reach maximum
t = 8.2 x 10⁻⁴ s
t = 0.82 ms
b) using law of conservation
I = 0.010 A
I = 10 mA
A car is traveling at 19m/s. It slows to a stop at a constant rate over 4.2 seconds. How far did the car travel during the 4.2se
1 answer:
The distance covered by the car is 39.9 m
Explanation:
Since the car is slowing at constant rate, it means that its acceleration is constant, therefore we can solve the problem by using the following suvat equation:
where
s is the distance covered
u is the initial velocity
v is the final velocity
t is the time elapsed
For the car in this problem, we have
u = 19 m/s (initial velocity)
v = 0 (final velocity)
t = 4.2 s (time elapsed)
Therefore, the distance covered is
Learn more about accelerated motion:
#LearnwithBrainly
You might be interested in
Answer:
2.30 × 10⁻⁸ N if the two electrons are in a vacuum.
Explanation:
The Coulomb's Law gives the size of the electrostatic force between two charged objects:
,
where
is coulomb's constant.
in vacuum.
and
are the signed charge of the objects.
is the distance between the two objects.
For the two electrons:
.
.
.
The sign of is negative. In other words, the two electrons repel each other since the signs of their charges are the same.