Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Answer:
4.4 cm
Explanation:
Given:
Distance of the screen from the slit, D = 1 m
Distance between two third order interference minimas, x = 22 cm
Let's say, minima occurs at:

We have:

Calculating further for the width of the central bright fringe, we have:

= 4.4 cm
Note: w in representswavelength
Answer:
8100W
Explanation:
Let g = 10m/s2
As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is
or 9000W
Since 10% of this is lost to friction, we take the remaining 90 %
P = 9000*90% = 8100 W