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3 years ago

How we can best share the observation of the light through one hole the paper

Physics

1 answer:

zhenek [66]3 years ago

5 0

Drawing is an easy way to connect to your inner child

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3. Waxing means "growing." Waning means

Sonja [21]

Answer:

I think it is the last one.

Explanation:

I am not sure because i am stuck on this one, too.

4 0

3 years ago

Read 2 more answers

As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

½ m v² = e ( $k \frac{Ze}{r^2}$ )

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

with this expression we can find the closest approach distance (r)

3 0

3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

How to represent milligram in kilogram by standard formula?

Anettt [7]

Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

we take $\frac{1}{1,000,000}$ which equals 0.000001 kg

4 0

3 years ago

Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the

Yuri [45]

Ans: Let d is the distance from height to our eyes.

<span>Applying the Pythagoras theorem, we get,</span>

<span>Check image: </span>

7 0

3 years ago