When light crosses the interface between a medium with higher refractive index and a medium with lower refractive index, there is a maximum value of the angle of incidence after which there is no refraction, but all the light is reflected, and this maximum value is called critical angle.
The critical angle is given by
where n1 is the refractive index of the first medium while n2 is the refractive index of the second medium. In our problem, n1=1.33 and n2=1.58, so the critical angle is
Answer:
8.362m/s
Explanation:
Given data
Mass m1= 7.77kg
Velocity v1= 7.77m/s
Mass m2= 8.88kg
Velocity v2= 8.88m/s
Apply the law of conservation of momentum for inelastic collision we have
m1v1+m2v2= (m+m2)V
7.77*7.77+ 8.88*8.88= (7.77+8.88)V
60.3729+78.8544= 16.65V
139.2273= 16.65V
Divide both sides by 16.65
V= 139.2273/16.65
V= 8.362m/s
Hence the final velocity is 8.362m/s
Answer:
Bill's motor power: W_B = F x S / T = F x 0.35 / 2= 0.175F
Nageen's motor power: W_N = F x S / T = F x 0.35 / 1.8 = 0.194F
=> 0.194F > 0.175F => Nageen's motor applied more power to the box than Bill's motor.
Answer:
The angle of deflection will be "1.07 × 10⁻⁷°".
Explanation:
The given values are:
Mass of a cow,
m = 1100 kg
Mass of bob,
mb = 1 kg
The total distance between a cow and bob will be,
d = 2 m
Let,
The tension be "t".
The angle with the verticles be "
".
Now,
Vertically equating forces
⇒ 
...(equation 1)
Horizontally equating forces
⇒ 
...(equation 2)
From equation 1 and equation 2, we get
⇒ 
O putting the estimated values, will be
⇒ 
⇒ 
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