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3 years ago

8

As the moon pictured revolves around the earth, the earth also rotates. how long is a lunar cycle?

1 answer:

Volgvan3 years ago

3 0

We don't know what you mean by "lunar cycle".

It appears at the same place in your sky every 24 hours and 50 minutes on the average.

It revolved around the Earth every 27.32 days.

It displays a full set of "phases" every 29.53 days.

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No file link please

musickatia [10]

Rock 3, because it has the most mass while still moving at the same speed, it is kinetic because there is movement and energy being used

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2 years ago

Jocelyn is running for the U.S Senate. What is one of the qualifications for office she must meet?

VMariaS [17]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

A proton is confined to a space 1 fm wide (about the size of an atomic nucleus). What's the minimum uncertainty in its velocity?

Daniel [21]

Answer:

Minimum uncertainty in velocity of a proton, $\Delta v\ge 3.15\times 10^7\ m/s$

Explanation:

It is given that,

A proton is confined to a space 1 fm wide, $\Delta x=10^{-15}\ m$

We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,

$\Delta p.\Delta x\ge \dfrac{h}{4\pi}$

Since, p = mv

$\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}$

$m \Delta v.\Delta x\ge \dfrac{h}{4\pi}$

$\Delta v\ge \dfrac{h}{4\pi m\Delta x}$

$\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}$

$\Delta v\ge 3.15\times 10^7\ m/s$

So, the minimum uncertainty in its velocity is greater than $3.15\times 10^7\ m/s$ . Hence, this is the required solution.

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3 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

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Damm [24]

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3 years ago