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3 years ago

6. how do the number of electrons flowing into a light bulb compare to the number of electrons flowing out of it?

1 answer:

N76 [4]3 years ago

5 0

If there's no leak in the bottom of the light bulb where electrons can
fall out onto the floor, and no magic electron generator inside the bulb
to create electrons out of thin air, then the number flowing in must be
the same as the number flowing out.

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How does a machine, such as a ramp, help make work easier?

Dovator [93]

Answer:

A machine can help decrease the input force and increase the output force.

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2 years ago

Determine the slit width that produces a diffraction pattern with the 2nd dark fringe at 6.2mm from the central fringe. The scre

Elanso [62]

Answer:

d= 0.242 mm

Explanation:

Slit width (d ) = ?

Screen distance ( D ) = 1.25 m

Wave length of light λ = 600 nm

Distance of n the dark fringe from centre

= n λ D / d

Here n = 2

$6.2\times10^{-3}=\frac{2\times600\times10^{-9}}{d}$

$d=\frac{1500\times10^{-6}}{6.2}$

d= 0.242 mm

4 0

3 years ago

Please answer this im not sure

vovikov84 [41]

Answer:

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3 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$