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mina [271]
3 years ago
15

Two objects, Object A and Object B, need to be identified. Object A's index of refraction is determined to be 1.77, and Object B

's index of refraction is determined to be 1.333. Knowing this information, which of the following must be true? A. Light can pass through Object A faster than it can pass through Object B.
B. The optical density of Object A is lower than the optical density of Object B.

C. Light can pass through Object B faster than it can pass through Object A.

D. The optical density of Object B is higher than the optical density of Object A.
Physics
2 answers:
Viefleur [7K]3 years ago
6 0

Penn Foster Students:  Light can pass through Object B  faster than it can pass through Object A

egoroff_w [7]3 years ago
6 0

C. Light can pass through Object B faster than it can pass through Object A.

Explanation:

The index of refraction of a material is given by

n=\frac{c}{v}

where

c is the speed of light in vacuum

v is the speed of light inside the material

From the equation, we see that there is an inverse relationship between n (the index of refraction) and v (the speed of light in the material): therefore, the lower the index of refraction, the higher the speed of the light.

In this problem, object A has a refractive index of 1.77, while object B has a refractive index of 1.333: since object B has a lower refractive index, light travels faster in object B.

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an object with a mass of 6 kilograms accelerates 4.0 MS to the second when an unknown force is applied to it what is the amount
SVEN [57.7K]
     Using the Newton's Secound Law, we have:

F=ma \\ F=6*4 \\ \boxed {F=24N}
7 0
3 years ago
How much energy becomes unavailable for work in an isothermal process at 440 k, if the entropy increase is 25 j/k?
just olya [345]
Answer: 11,000 J

Explanation:

In an isothermal process,

\text{entropy increase} =  \frac{\text{amount of energy in heat transfer}}{\text{temperature} }  (1)

Note that, the energy used in heat transfer is not available for work. So, the amount of energy unavailable for work is equal to the energy used in heat transfer.

To obtain the amount of energy in heat transfer, we multiply both sides of equation (1) by the denominator of the right side of (1) so that 

amount of energy in heat transfer = (entropy increase)(temperature)
                                                      = (25 J/K)(440 K)
                                                      = 11,000 J

Since the amount of energy unavailable for work is equal to the amount of energy in the heat transfer, therefore the amount of energy unavailable for work is 11,000 J.
8 0
3 years ago
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h
-Dominant- [34]

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

       L₀ = L_{f}

       I₁ w₁ = I₂ w₂

       w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

       w₂ = 22.6 rad/s

3 0
3 years ago
You dip your finger into a pan of water twice each second, producing waves with crests that are separated by 0.15 m. Determine t
PIT_PIT [208]

Frequency = rate of sploosh = 2 per second  =  2 Hz.

Period = ( 1/frequency ) =  1/2  second

Speed = (wavelength) x (frequency) = (0.15m) x ( 2/sec) = 0.075 m/s .

5 0
3 years ago
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi
morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = \dfrac{0.029158}{4.03176}= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

4 0
3 years ago
Read 2 more answers
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