components of the speed of the coin is given as




now the time taken by the coin to reach the plate is given by



now in order to find the height



so it is placed at 1.52 m height
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:
option (B)
Explanation:
Intensity of unpolarised light, I = 25 W/m^2
When it passes from first polarisr, the intensity of light becomes

Let the intensity of light as it passes from second polariser is I''.
According to the law of Malus

Where, θ be the angle between the axis first polariser and the second polariser.

I'' = 11.66 W/m^2
I'' = 11.7 W/m^2
Answer:
distance between both oasis ( 1 and 2) is 27.83 Km
Explanation:
let d is the distance between oasis1 and oasis 2
from figure
OC = 25cos 30
OE = 25sin30
OE = CD
Therefore BC = 30-25sin30
distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem
in

PUTTING ALL VALUE IN ABOVE EQUATION


d = 27.83 Km
distance between both oasis ( 1 and 2) is 27.83 Km