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4 years ago

Which best describes what happens to sunlight after passing through the keyhole of a door

1 answer:

7 0

Given the answers I would say B, Is your best choice because when the light goes through a hole it spreads out, because most light rays (<) outward.

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Light traveling through the air strikes the surface of the four materials below. Which material reflects light but does not refr

aliina [53]

I could of sworn it was d metal spoon

7 0

3 years ago

Read 2 more answers

A 43-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d

lesya [120]

Answer:

$d_{2}=0.585m$

Explanation:

The torque is the force by the distance so to determinate that both torque are the same magnitude so

$T_{1}-T_{2}=0$

$T_{1}=T_{2}$

$F_{1}*d_{1}=F_{2}*d_{2}$

$43N*0.30m=21N*d_{2}$

Solve to d2

$d_{2}=\frac{41N*0.30m}{21N}$

$d_{2}=0.585m$

5 0

3 years ago

Fermi energy of conduction electrons in silver is 0.548 J. Calculate the number of such electrons in unit cm3

ICE Princess25 [194]

Answer:

$n=1.86*10^{-30}m^3$

Explanation:

From the question we are told that:

Fermi energy of conduction electrons $E_f=0.548J$

Generally the equation for Fermi energy is mathematically given by

$E_f=\frac{3}{\pi}^2/3*\frac{h^2}{8m}*{n^{2/3}}$

${n^{2/3}}=\frac{E_f}{\frac{3}{\pi}^2/3*\frac{h^2}{8m}}$

Where

h= Planck's constant

${n^{2/3}}=\frac{E_f}{\frac{3}{\pi}^2/3*\frac{h^2}{8m}}$

${n^{2/3}}=\frac{0.548J}{3.62*10^{-19}}$

$n=(1.51*10^{-20})^{3/2}$

$n=1.86*10^{-30}m^3$

3 0

3 years ago

A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second

Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$

Thus, the value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0

2 years ago

Relate the output of energy from a heat engine to the energy put into the heat engine considering the second law of thermodynami

Vinil7 [7]

<u>Answer: </u>

<em>Considering the II law of thermodynamics</em>

<em>From the figure</em>

<em>Out put of energy: </em>

Heat supplied from the source/ reservoir (Q₁) - Heat rejected to the surroundings from the system (Q) = Q₁ - Q₂. Also known as Net work done on the system.

<em>Input of energy: </em>

Amount of heat energy supplied to the system from the source (Q₁ ).

Efficiency (H.E) = η = Output÷ Input

η = (Q₁ - Q₂) ÷ Q₁

OR η = Wnet ÷ Q₁ ; since Wnet = (Q₁ - Q₂)

3 0

3 years ago