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ss7ja [257]
3 years ago
13

Students have 4 different pairs of sunglasses, each with lenses where the polarization filter is oriented in different direction

s (indicated by the direction of lines drawn on the lenses below). They are planning to ride rafts down a river on a sunny day where there is lots of glare on the water. Which pair would they want to wear to eliminate the glare from the sun
Physics
1 answer:
AysviL [449]3 years ago
6 0

Answer:

the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.

Explanation:

The glare is caused by the reflection of light in the water, the polarization of the reflected light is polarized in a direction parallel to the surface of the water, the polarization is total for the angles

         n = tan  \theta_{p}

         \theta_{p} = tan⁻¹ n

the refractive index for seawater is 1.33

         \theta_{p}= tan⁻¹  1.33

         \theta_{p} = 53º

for this angle the light is totally polarized, for the other angles the polarization is partial.

Based on this, the lenses must eliminate this polarization, so its polarization direction must have 90º with respect to this polarization,

          \theta_{lens} = 53 +90

           \theta_{lens}= 143º

Therefore, the lens you must select has an angle of 143º measured with respect to the horizontal, this angle is 53º with respect to the vertical.

A lens that could work is one that is polarized 45º with respect to the vertical.

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3 ways how natural disasters impact environment.
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2 years ago
A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0
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