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3 years ago

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A ray of light is traveling in a glass cube that is totally immersed in water. you find that if the ray is incident on the glass

– water interface at an angle to the normal larger than 48.7°, no light is refracted into the water. what is the refractive index of the glass?

1 answer:

mihalych1998 [28]3 years ago

6 0

I'm sorry I really wished I could answer.

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Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,

Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41 , RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc = 364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -) √ 150.74² 77.0²

RL = 473 (+ / -) (129.58)

RL₁ = 34.41 , RL₂ = 602.58

d.

The higher value have the less angular frequency

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0

3 years ago

In what state must matter exist for fusion reactions to take place?<br><br> ...?

valina [46]

It must exist in p<span>lasma state.</span>

4 0

3 years ago

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List five situations in which a lesser amount of friction would be beneficial. Overall, would a high or low coefficient of frict

Natalka [10]

Low coefficient of friction

1. flying a plane (friction between air and plane)
2. ice skating (friction between ice and skate blade)
3. swimming (water & skin)
4. rowing a boat (water and boat)

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3 years ago

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Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

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3 years ago

Objects 1 and 2 attract each other with a gravitational force of 45 units. If the mass of Object 1 is doubled, then the new grav

Novosadov [1.4K]

Explanation:

Fgravity = G*(mass1*mass2)/D²

so, if you double one of the masses, what does that do to our equation ?

Fgravitynew = G*(2*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity

so, the correct answer will be 2×45 = 90 units.

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2 years ago

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