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Helen [10]
2 years ago
11

A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an

angle of 23.0° with the normal. What is the index of refraction of the plastic?
Physics
1 answer:
Cloud [144]2 years ago
5 0

Answer:

1.36

Explanation:

n_{air} = Index of refraction of air = 1

n_{plastic} = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

n_{air} Sini = n_{plastic} Sinr

(1) SIn32.0 = n_{plastic} Sin23.0

n_{plastic} = 1.36

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A monoatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.010 m^3 to 0.040 m^3. The fina
Natali [406]

Answer:

A) 0.0 kJ

Explanation:

Change in the internal energy of the gas is a state function

which means it will not depends on the process but it will depends on the initial and final state

Also we know that internal energy is a function of temperature only

so here the process is given as isothermal process in which temperature will remain constant always

here we know that

\Delta U = \frac{3}{2}nR\Delta T

now for isothermal process since temperature change is zero

so change in internal energy must be ZERO

4 0
2 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
The graph below shows the sunspot number observed between 1750 and 2000. The graph shows sunspot number on the y axis and years
OlgaM077 [116]
1850 to 1900 because the slope would be 105. It says what is the greatest fall, so the upward slope of 120 wouldn't count.
3 0
2 years ago
Read 2 more answers
Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to
weqwewe [10]
Arginine is a basic aminoacid, because it has two amino groups and one acid group. At a low pH, every ionizable group is protoned. At a little higher pH, the acid group looses its proton. A little higher pH, one amino group looses its proton. At a very high pH, all ionizable groups are not protoned.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> So 9.20 is higher tan the second pKa and lower than the third pka. This means the acid has already lost its proton, and one of the aminos too, but the second amino hasn’t. When an acid is not protoned, it has a negative charge. When an amino is not protoned, it’s neutral. When an amino is protoned, it has a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative charge (the acid), what makes it neutral.
4 0
2 years ago
Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. D
Ymorist [56]

Answer : The rate of heat transfer to the water is, 37.92 kJ/min

Explanation : Given,

Time = 10 min

Mass of water = 200 g

Latent heat of fusion of water = 334 J/g

Latent heat of vaporization of water = 2230 J/g

Now we have to calculate the rate of heat transfer to the water.

Q=\frac{m\times (L_v-L_f)}{t}

Now put all the given values in the above formula, we get:

Q=\frac{200g\times (2230-334)J/g}{10min}

Q=37920J/min=37.92kJ/min

Thus, the rate of heat transfer to the water is, 37.92 kJ/min

4 0
2 years ago
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