Question

I will pick the branliest if answered these questions.


1. A light wave moves from diamond (n=2.4) into water (n=1.33) at an angle of 24 degrees. What angle does it have in water?

2.Given that the specific heat or water is 4.18KJ/(Kg°C), how much heat does it take to raise the temperature of 3.5 kg of water from 25°C to 55°C?

3.What is the average velocity of atoms in 1.00 mol of argon (a monatomic gas) at 275K? For m, use 0.0399 kg.

4.What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?

5. A force of 3300 N is exerted on a piston that has an area of 0.060m^2. What force is exerted on a second piston that has an area of 0.18 m^2?

6.What is the equivalent resistance if you connect three 10.0 resistors in series?

7.What is the equivalent resistance if you connect three 10.0 resistors in parallel?

8. Calculate the current in each 10.00 resistor in the series circuit and in the parallel circuit if the power supply is 60.0 V.


Thank you.

Answer 1

1.

n₁ = index of refraction of diamond = 2.4

n₂ = index of refraction of water = 1.33

θ₁ = angle of incidence = 24 deg

θ₂ = angle of refraction = ?

using snell's law

n₁ Sinθ₁ = n₂ Sinθ₂

inserting the values

(2.4) Sin24 = (1.33) Sinθ₂

θ₂ = 47.22 deg

2.

c = specific heat = 4.18

m = mass of water = 3.5 kg

ΔT = change in temperature = 55 - 25 = 30 C

Q = heat taken

heat taken is given as

Q = m c ΔT

inserting the values

Q = (3.5) (4.18) (30)

Q = 439 J

3.

n = number of moles = 1

m = molar mass = 0.0399 kg

T = temperature = 27 K

v = average velocity

average velocity is given as

v = sqrt(3RT/m)

v = sqrt(3 x 8.314 x 275/0.0399)

v = 414.6m/s

4.

n = number of moles = 2

T = temperature = 35 C  = 35 + 273 K = 308 K

U = internal energy

internal energy is given as

U = 2.5 n RT

U = 2.5 (2) (8.314) (308

U = 1.28 × 10⁴ J

5.

F₁ = 3300 N

F₂ = ?

A₁ = 0.060 m²

A₂ = 0.18 m²

Using pascal's law

F₁/A₁ = F₂/A₂

3300/0.060 = F₂/0.18

F₂ = 9900 N

6.

R = resistance of each resistance in series = 10 ohm

R' = equivalent resistance in series

equivalent resistance is given as

R' = 3 R

R' = 3 x 10

R' = 30 ohm

7.

R = resistance of each resistance in parallel = 10 ohm

R'' = equivalent resistance in parallel

equivalent resistance is given as

R'' = R/3

R'' = 10/3

R'' = 3.3 ohm

8.

for series circuit

R' = equivalent resistance in series = 30 ohm

V = Voltage applied = 60 Volts

i' = current in each resistor in series

current in each resistor in series is given as

I' = V/R'

i' = 60/30

i' = 2 A

for parallel circuit :

i = current in each resistor = V/R = 60/10 = 6 A

Answer 2

Answer:

1) 46.88°

2) 438,900Joules

3) 414.62m/s

4) 1.28×10⁴Joules

5) 9900N

6) 30ohms

7) 3.33ohms

8) current in each 10ohms resistor in series is 2A.

current in each 10ohms resistor in parallel is 6A

Explanation:

Find the solutions to questions 1 to 7 in the attachment below.

8) If the resistance in series are connected to a 60volts supply, the current in each resistor is calculated using ohms law.

E = IRt

E is the supply voltage = 60V

Rt is the total equivalent resistance = 3(10) = 30ohms

I is the total current

I = E/Rt

I = 60/30

I = 2A

Note that in a series connected circuit, same current flies through the resistors, therefore 2A will be the current in each 10ohms resistors

FOR PARALLEL CONNECTION, same voltage but different current flows through the resistors.

Using V = IR

V = 60V

I = I1 = I2 = I3 = V/R

= 60/10

= 6A

Note that for parallel connection, the individual resistances are used to get the current instead of their equivalent resistance as used in series connection.