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netineya [11]
3 years ago
7

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1515

.
Physics
1 answer:
Talja [164]3 years ago
8 0

Complete  Question

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1/5.

Answer:

The  angle is  

Explanation:

From the question we are told that

   The light emerging from second Polaroid is 1/5 the  unpolarized

Generally the intensity of light emerging from the first Polaroid is mathematically represented as

             I_1 = \frac{I_o}{ 2 }

Generally from the Malus law the intensity of light emerging from the second Polaroid  is mathematically represented

      I_2  =  I_1 cos^2 (\theta )

=>   cos^2 (\theta ) =  \frac{I_2}{I_1 }

=>   cos (\theta) =  \sqrt{ \frac{I_2}{I_1} }

From the question I_2  =  \frac{I_o}{5}

     cos (\theta) =  \sqrt{ \frac{ \frac{ I_o}{5} }{\frac{I_o}{2} } }

     cos (\theta) =  \sqrt{ \frac{2}{5} }

=>    \theta =   cos ^{-1} [\sqrt{\frac{2}{5}}  ]

=>    \theta  =  50.77^o

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A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i
padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

v=\sqrt{\dfrac{T}{\mu}}

Where, \mu = mass per unit length

T = tension

Put the value into the formula

v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}

v =519.61\ m/s

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0
3 years ago
A student examines a 20-meter long rectangular stream channel and takes the following measurements: width of stream = 4 meters,
Scrat [10]

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.

3 0
3 years ago
The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.
pickupchik [31]

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × 10^{-6} = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

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As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

w_{f} = w_{i} + α × t

Where

w_{f} is the final angular velocity

w_{i} is the initial angular velocity

α is the angular acceleration

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w_{i} is 0 (∵ initially it starts from rest)

By substituting the values we get

w_{f} = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

3 0
3 years ago
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