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3 years ago

A light ray propagates from air to glass. The refractive index for the glass is....

Physics

2 answers:

natita [175]3 years ago

4 0

Answer:

1.5

Explanation:

sin i / sin r = n

i = 135 - 90 = 45

r = 118 - 90 = 28

sin (45) / sin (28) = n

n= 1.5

motikmotik3 years ago

3 0

Answer:

118 to glass

Explanation:

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If a car goes along a straight road heading east and speeds up from 45ft/s to 60ft/s in 5s , calculate the acceleration

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Acceleration = change in velocity/ time = (final velocity - initial velocity)/time

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4 years ago

An object is travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the

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50 = 0 + 0(4) + ½a(4²)

50 = 8a

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3 years ago

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

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3 years ago

A CG amplifier using an NMOS transistor for which gm = 2 mA/V has a 5-k drain resistance RD and a 5-k load resistance RL. The am

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Answer: Input Resistance (Rin) = 500 ohms , overall voltage Gain (GV) = 2 volts / volts , ID = 4/9 = 0.44 A

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5 0

3 years ago

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

QveST [7]

Answer:

$\Delta \theta = 47.57^{\circ} C$

Explanation:

given,

moles of air compressed, n = 1.70 mol

initial temperature, T₁ = 390 K

Power supply by the compressor, P = 7.5 kW

Heat removed = 1.3 kW

Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.

Time of compression = time of the hay revolution

= $\dfrac{1}{2}\ T$

= $\dfrac{1}{2}\times \dfrac{1}{f}$

= $\dfrac{1}{2}\times \dfrac{1}{1.833}$

=0.273 s

Using first law of thermodynamics

U = Q - W

now,

$\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}$

Power supplied $\dfrac{\Delta W}{\Delta t}$ = 7.5 kW

heat removed $\dfrac{\Delta Q}{\Delta t}$ = 1.3 kW

now,

$\dfrac{\Delta U}{\Delta t} = 7.5 -1.3$

$\dfrac{\Delta U}{\Delta t} = 6.2 kW$

we know,

$\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}$

C_v for air = 5 cal/° mol

= 5 x 4.186 J/mol°C = 20.93 J/mol°C

now,

$\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}$

$\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}$

$\Delta \theta = 47.57^{\circ} C$

the temperature change per compression stroke is equal to 47.57°C.

4 0

3 years ago