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jek_recluse [69]
2 years ago
4

Explain why it is not advisable to use small values of incident ray in performing experiment on the refraction through a glass p

rism.
Physics
2 answers:
KATRIN_1 [288]2 years ago
8 0
Probably to do with the geometry behind snell's laws of refraction. in a prism you want a measurable deviation, and that can I think be helped by having a large angle of incidence at the first face of the prism
QveST [7]2 years ago
7 0

Answer:

As we know that when light incident on a glass prism then few amount of the light is absorbed by the prism and then the remaining amount of light gets refracted out of the prism on the other side.

In this whole experiment of the prism we know that the amount of light absorbed by the prism completely depends on the material and in order to find the phenomenon of refraction through prism we required a good amount of light to be incident on the prism.

Because if we took small amount of light the the refracted light that comes out of prism is very small to detect the complete phenomenon of refraction.

This is the reason why it is not advisable to use small values of incident ray in performing experiment on the refraction through a glass prism.

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pogonyaev
If you are asking what the volume of the cube is it would be 20.3 - 17.5 ml so 2.8 ml.
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2 years ago
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A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme
GaryK [48]

Given that,

Distance in south-west direction = 250 km

Projected angle to east = 60°

East component = ?

since,

cos ∅ = base/hypotenuse

base= hyp * cos ∅

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East component = 125 km

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A runner has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?
antiseptic1488 [7]
I could be wrong, but I'm pretty sure it's 144kg.
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An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10
Marianna [84]

Answer:

a = 18.28 ft/s²

Explanation:

given,

time of force application, t= 10 s

Work = 10 Btu

mass of the object = 15 lb

acceleration, a =  ? ft/s²

1 btu = 778.15 ft.lbf

10 btu = 7781.5 ft.lbf

m = \dfrac{15}{32.174}\ slug

m = 0.466 slug

now,

work done  is equal to change in kinetic energy

W = \dfrac{1}{2} m (v_f^2-v_i^2)

7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2

 v_f = 182.75\ ft/s

now, acceleration of object

  a = \dfrac{v_f-v_o}{t}

  a = \dfrac{182.75-0}{10}

         a = 18.28 ft/s²

constant acceleration of the object is equal to 18.28 ft/s²

3 0
2 years ago
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0
2 years ago
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