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3 years ago

The speed of light in a material is 0.50

Physics

1 answer:

vivado [14]3 years ago

4 0

Answer:

Explanation:

This problem indicates that the speed of light in a material medium is 0.5 10⁸ m / s, they ask to find the critical angle between the material and the vacuum

Let's find the refractive index of the material

n = c / v

n = 3 10⁸ / 0.5 10⁸

n = 6

When the material passes from one medium to another, it must comply with the law of refraction

n₁ sin θ = n₂ sin θ₂

for the angle criticize the angles tea2 = 90

tea = sin⁻¹n₂/ n₁

The vacuum replacement index is n₂ = 1

tea = sin⁻¹ (1/6)

tea = 9.59º

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Which item(s) would be sufficient to make a circuit?

masya89 [10]

Explanation :

A circuit is the representation of the path of the flow of current. The circuit can be either closed or open.

When the switch is off the circuit is closed circuit and when the switch is not connected the circuit is open.

The items that are sufficient to make a circuit are as follows :

Voltage source like a battery.
Resistors or electrical equipment like heater, motor etc.

Other components can be ammeter, voltmeter, ac source, variable resistors etc.

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3 years ago

Read 2 more answers

A certain heat engine takes in 300 J of energy from a hot source and then transfers 200 J of that energy to a colder object. Wha

Greeley [361]

Answer:

The efficiency is 0.33, or 33%.

Explanation:

From the thermodynamics equations, we know that the formula for the efficiency of a heat engine is:

$\eta=1-\frac{Q_2}{Q_1}$

Where η is the efficiency of the engine, Q_1 is the heat energy taken from the hot source and Q_2 is the heat energy given to the cold object. So, plugging the given values in the formula, we obtain:

$\eta=1-\frac{200J}{300J}=0.33$

This means that the efficiency of the heat engine is 0.33, or 33% (The efficiency of an engine is dimensionless).

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3 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

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2 years ago

A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of

enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}$

In the old transformer the ratio of voltages was:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1$

In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$

If the old current output was 425 kV, the ratio of current was:

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1$

Then, the ratio of the new output over the old output is:

$\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86$

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

$P_L=I^2\cdot R$

Then, the ratio of losses is (R is constant for both power losses):

$\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74$

The losses in the new line are 74% the losses in the old line.

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3 years ago

MAJOR POINTS! Solve the system of equations to find <img src="https://tex.z-dn.net/?f=%20a_%7B2%7D%20%20" id="TexFormula1" title

Monica [59]

You are dealing with pulleys?

can be done with addition of the two equations to eliminate T.

$(2^{*}) \quad {} -m_1 g \cos \theta + T = m_1 a _2$
+
$(4^{*}) \quad 3m_1 g - T = 3m_1 a_2$
=

$(-m_1 g \cos\theta + 3m_1 g) + (T - T) = m_1 a _2 + 3m_1 a_2 \implies \\ \\
(-m_1 g \cos\theta + 3m_1 g) = 4m_1 a_2 \implies \\ \\
m_1(- g \cos\theta + 3g)= 4m_1 a_2$

we can cancel m₁ by dividing both sides by it, assuming mass is not zero

$(- g \cos\theta + 3g)= 4a_2 \implies \\ \\
a_2 = \dfrac{- g \cos\theta + 3g}{4} \\ \\
a_2 = \dfrac{- 9.80 \cos 60 + 3(9.80)}{4} \\ \\
a_2 = 6.125 \text{m/s}^2
$

a₂ = 6.125 m/s² ( do significant digits if you need to)

6 0

3 years ago