Answer:




Explanation:
The electrical reactance is defined as:

Where:

So, replacing the data provided by the problem:

Now, the impedance can be calculated as:

Where:

Replacing the data:

In order to find the magnitude of the impedance we can use the next equation:

We can use Ohm's law to find the current:

Therefore the current is:

And its magnitude is:

Finally the phase angle of the current is given by:

Answer:
Answer:
4 ms
Explanation:
initial velocity, u = 75 m/s
final velocity, v = 0
distance, s = 15 cm = 0.15 m
Let the acceleration is a and the time taken is t.
Use third equation of motion
v² = u² + 2 a s
0 = 75 x 75 - 2 a x 0.15
a = - 18750 m/s^2
Use first equation of motion
v = u + at
0 = 75 - 18750 x t
t = 4 x 10^-3 s
t = 4 ms
thus, the time taken is 4 ms.
Explanation:
Answer:
your velocity is 2.5 m/sec