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Mila [183]
3 years ago
14

I will pick the branliest if answered these questions.

Physics
2 answers:
Bond [772]3 years ago
6 0

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








laiz [17]3 years ago
5 0

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.

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A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ
DochEvi [55]

Answer:

a)  k=19.6N/m

b)  V_m=0.81m/s

c)  a_m=6.561m/s^2

d)  K.E=0.096J

e)  T=0.78sec &F=1.29sec

f)   mx'' + kx' =0

Explanation:

From the question we are told that:

Stretch Length L=0.150m

Mass m=0.30kg

Total stretch lengthL_t=0.150+0.100=>0.25

a)

Generally the equation for Force F on the spring is mathematically given by

F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}

k=19.6N/m

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

V_m=A\omega

Where

A=Amplitude

A=0.100m

And

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Therefore

V_m=A\omega\\\\V_m=8.1*0.1

V_m=0.81m/s

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

a_m=\omega^2A

a_m=8.1^2*0.1

a_m=6.561m/s^2

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*0.3*0.8^2

K.E=0.096J

e)

Generally the equation for  the period T is mathematically given by

\omega=\frac{2\pi}{T}

T=\frac{2*3.142}{8.1}

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Generally the equation for  the Frequency is mathematically given by

F=\frac{1}{T}

F=1.29sec

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

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Where

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7 0
3 years ago
If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
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Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
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