Explanation:
The given data is as follows.
Velocity of bullet,
= 814.8 m/s
Observer distance from marksman, d = 24.7 m
Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.
t =
(velocity in air = 343 m/s)
= 0.072 sec
Now, before the observer hears the report the distance traveled by the bullet is as follows.

= 
= 58.66
= 59 (approx)
Thus, we can conclude that each bullet will travel a distance of 59 m.
Answer:
false
Explanation:
It doesn't the copper wire wouldn't even be pulled by the magnet at all and the electricity would stay inside of the the force of the copper wire
Answer: 7.53 μC
Explanation: In order to explain this problem we have to use the gaussian law so we have:
∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0
Q inside= 0 = q+ Qinner surface=0
Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC
Explanation:
It is given that,
Bandwidth of a laser source, 
(b) Let t is the time separation of sections of sections of the light wave that can still interfere. The time period is given by :



(a) Let h is the coherence length of the source. It is given by :

c is the speed of light

l = 0.0099 m
Hence, this is the required solution.
Answer:
a)
b)
c)
d)
e)
Explanation:
Given that
d = 2 cm
V = 200 V

We know that
F = E q
F = m a
E = V/d
So
m a = q .V/d b
---------1
The mass of electron

The charge on electron

Now by putting the all values in equation 1


We know that
a)
s = 0.1 cm


b)
s = 0.5 cm


c)
s = 1 cm


d)
s = 1.5 cm


e)
s = 2 cm

