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Liula [17]
3 years ago
12

Approximately what percent of the Moon’s surface is illuminated on a new-moon day? A) 0% B) 20% C) 50% D) 100%

Physics
2 answers:
fomenos3 years ago
4 0
50% (half) of the moon's surface is illuminated at all times. At the time of New Moon, because of the moon's position between us and the sun, the illuminated half is turned completely away from us, and we can't see any of it.
Sergeeva-Olga [200]3 years ago
3 0
At New Moon the percent illuminated is 0; at First and Last Quarters it is 50%; and at Full Moon it is 100%<span>. During the crescent phases the percent illuminated is between 0 and </span>50%<span> and during gibbous phases it is between </span>50%<span> and </span>100%<span>.
so it depends on the phase 


</span>
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Which statement is true? Which statement is true? An orbital that penetrates into the region occupied by core electrons is more
masya89 [10]

Hey there!:

Here the Statement - D is correct.  

Because Orbitals containing the core electrons are more attracted towards nuclear charge and hence less shilded from nuclear charge than an orbital that doesn't penetrate.  Also due to more attraction between the orbital containing core electron and nucleus, it will have less energy.

Hope this helps!

6 0
3 years ago
How many moons does Venus have?
Gennadij [26K]
The answer is no moons<span> at all. That's right, </span>Venus<span> (and the planet Mercury) are the only two planets that don't </span>have<span>a single natural </span>moon<span> orbiting them. Figuring out why is one question keeping astronomers busy as they study the Solar System.</span>
4 0
3 years ago
Read 2 more answers
PLS HELP!! YOU CAN SKIP THE INFO IF YOU WANT!!
agasfer [191]

Answer:

B

Explanation:

The whole thing is talking about the damage runoffs have done that is equal to answer B.

5 0
2 years ago
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Sketch a reaction progress curve for a reaction that has an activation energy of 22 kj and the total energy change is -103kj.
crimeas [40]

Answer:

Do find the answer in the attachment herein.

Explanation:

From the attached diagram:

I. Activation energy = Activated complex - ∆H(reactants)

Activation energy = 162-140 = 22Kj.

II. ∆H(reaction) = ∆H(products) - ∆H(reactants)

∆H(reaction) = 37 - 140 = -103Kj.

8 0
4 years ago
You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a
gtnhenbr [62]

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}

Where;

d is the separation or distance between the two parallel plates;

d = \frac{VL^2 \epsilon_o}{Q} \\\\d =  \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

6 0
4 years ago
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