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3 years ago

Calculate the angle of incidence of light ray incident on surface

Physics

1 answer:

Katena32 [7]3 years ago

3 0

Answer:

$from \: snell {}^{.} s \: law : \\ n = \frac{ \sin(i) }{ \sin(r) } \\ \sqrt{3} = \frac{ \sin(i) }{ \sin(30 \degree) } \\ \sin(i) = \sqrt{3} \times \sin(30 \degree) \\ \sin(i) = 0.866 \\ i = { \sin}^{ -1 } (0.866) \\ i = 60 \degree$

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In one hour, coal supplies 500 000 J of energy. The energy amounts to 380 000 J. How much useful energy is produced in one hour?

Debora [2.8K]

Answer:

120,000J

Corrected question;

In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?

Explanation:

Given;

Total energy Et = 500,000 J

Wasted Energy Ew = 380,000J

The amount useful energy is the amount of energy that is available for supply.

This can be derived by subtracting the wasted energy from the total energy.

Useful energy = Total Energy - wasted energy

Eu = Et - Ew

Substituting the given values;

Eu = 500,000J - 380,000

Eu = 120,000 J

The amount of useful energy produced in one hour is 120,000 J

5 0

3 years ago

I want to answer these questions about vectors

stellarik [79]

Answer:

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3 years ago

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

3 years ago

A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.

Elza [17]

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant $k=4.45\times 10^3 N/m$

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

$P_i=\frac{kx^2}{2}$

$P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J$

$P_f=mgh=0.25\times 9.8\times h$

$P_i=P_f$

$14.24 =0.25\times 9.8\times h$

$h=\frac{14.24}{0.25\times 9.8}=5.81 m$

6 0

3 years ago

Why does data need to be reliable

Karolina [17]

So results can be shared and used by other scientists that want to use or replicate your experiment.

8 0

3 years ago

Read 2 more answers