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1 year ago

How does a 12-month lunar calendar differ from our 12-month solar calendar?.

Physics

1 answer:

horrorfan [7]1 year ago

8 0

Answer:

It has about 11 fewer days. It does not have seasons. Its new year always occurs in February instead of on January 1.

Explanation:

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A laser beam is incident on two slits with a separation of 0.215 mm, and a screen is placed 5.45 m from the slits. an interferen

vladimir1956 [14]

Use: dsin(θ) = mλ where d is slit separation, m is fringe order (1 here), and
θ = 0.183
Now λ = dsin(θ) /m = (0.215e-3)(sin(0.183))/1 = 6.867e-7 or λ = 687.7nm
(red laser)

4 0

2 years ago

Select the correct answer.

Morgarella [4.7K]

D all of them is the answer. All three can be used to celebrate sporting events

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1 year ago

Bicycle in its writers have combined a mass of 80 kg in a speed of 6.0 meters per Second what is the magnitude of the average fo

erik [133]

Answer:

F = 120 N

Explanation:

Force x distance = energy

The bike has energy 1/2 . 80 . 6^2 = 1440 J

You are looking at an example of not reading the question properly.

Impulse = Force . time = change in momentum

F . 4 = 80 .6

F = 120 N

7 0

2 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

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

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

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

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

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2 years ago

What planet has a visible mark referred to as the the Great Red Spot?

vivado [14]

Jupiter that is the answer

good luck

6 0

2 years ago