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4 years ago

8

The length of daylight on the moon is about __________.

2 answers:

insens350 [35]4 years ago

4 0

2 week of daylight and 2 weeks of night

Lyrx [107]4 years ago

3 0

<span>The length of daylight on the moon is about 29.5 days.</span>

From the earliest days, the Moon has been there in the Solar System and there has never been a period when we couldn't gaze upward in the night sky and either observe the Moon hanging there, or realize that it would be back the precise one night from now (i.e. a New Moon).

A day on the Moon keeps going as long as 29.5 Earth days. We can say that it would take 29.5 days for the Sun to move the distance over the sky and come back to its unique position once more.

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Witch unit is used to measure mass in the metric system

Alecsey [184]

The gram (g)
a nickel weighs 5 grams, and a raisin weighs approximately 1 gram.
There is also the kilogram(kg), or 1000 grams

6 0

3 years ago

Read 2 more answers

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

3 0

4 years ago

I need help plzzzz!!!!!!!!!!!!!!!!

elena-s [515]

Answer:

You were a freeloader of my questions, so I'll be one too.

4 0

3 years ago

I'm not an idiot so I'm pretty sure c and d are not the answers. I'm confused because don't they both do this. As it picks up sp

krok68 [10]

The answer is A because it's MOVING

ithink

8 0

3 years ago

Read 2 more answers

The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra

Scrat [10]

Answer:

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

Explanation:

Given

Number of protons $= 92$

Radius of nucleus $r_n = 7.4 * 10^{-15} m$

Distance of the electrons $r_1 = 1.0 * 10^ {-10} m$

Part 1

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21} N/C$

Part 2

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13} N/C$

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

7 0

3 years ago