To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.
From the law of Malus intensity can be defined as
Where
Angle From vertical of the axis of the polarizing filter
Intensity of the unpolarized light
The expression for the intensity of the light after passing through the first filter is given by
Replacing we have that
Re-arrange the equation,
Re-arrange to find \theta
The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°
Answer:
Explanation:
We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ
Applying law of conservation of momentum along direction of original motion
m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄
0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ
v cos θ = .8
Applying law of conservation of momentum along direction perpendicular to direction of original motion
1.03 sin 43 x .143 = .132 x v sinθ
v sinθ = .76
squaring and adding
v² = .76 ² + .8²
v = 1.1 m /s
Tan θ = .76 / .8
θ = 44°
Given :
Initial speed , u = 0 m/s .
Final speed , v = 91 km/h = 25.28 m/s .
To Find :
a) Average acceleration .
b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .
Solution :
a )
We know ,by equation of motion :
b)
Also , by equation of motion :
Hence , this is the required solution .
Answer:
The discharge of the stream at this location is 40 cubic meters per second.
Explanation:
The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:
Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)
Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)
Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)
<u>Volume Flow Rate = 40 cubic meters per second</u>
Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>
This result shows that 40 cubic meters volume of water passes or discharges through this point in a time of one second. Hence, this is called the volume flow rate or the discharge of the stream.
7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
