We know, R = V / I
Here, V = 86 V
I = 3 A
Substitute their values,
R = 86 / 3
R = 28.67 Ohm
In short, Your Answer would be 28.67 Ohms
Hope this helps!
Answer:
The answer to your question is 1.35 Watts
Explanation:
Data
Work = W = 5 J
time = t = 3.7 s
Power = P = ?
Formula
Power is a rate in which work is done or energy is transferred over time
P = 
Substitution

Result
P = 1.35 W
I wasn't there observing the experiment while you and your class
performed it, so I don't really know how it was set up, or what
happened.
But I can tell you this: Light doesn't bend while passing through
any medium. It only bends at the boundary where one medium
meets a different one.
Please find the figure attached below:
Answer:
a.Resistance of bulb=
=192 ohm
b.power consumed by bulb after motor disconnection=74.609 W
Explanation:
<u>a.Resistance of bulb=</u>
<u>=?</u>
As we know that 
putting R as resistance of bulb i.e. 

<u>b.power consumed by bulb after motor disconnection? </u>
from the figure we see that
are in series so

current through their resistance is

Power consumed by bulb is
