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sattari [20]
3 years ago
13

HELP!!! ASP EASY QUESTION! WILL GIVE BRAINEST IF U ANSWER!

Physics
1 answer:
Harlamova29_29 [7]3 years ago
5 0
I believe the answer is Anther.
The Anther produces the pollen
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The gamma ray released by each decay carries 140kev of energy. find the total energy e released by decays in the 2 hours.
olya-2409 [2.1K]
First, we would need to know the decaying isotope.
Next, we use the decay formula
A = Ao e^(-kt)
After determining the remaining amount after two hours, the decay reaction can be used to determine the number of gamma rays released. If the given is in terms of mole, then the total energy is
E = 140n KeV where n is the number of moles of gamma rays released
8 0
3 years ago
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If you were to divide the presentday universe into cubes whose sides are 10 million lightyears long, each cube would contain, on
vovikov84 [41]

Answer:

Explanation:

density of galaxies would be \frac{1}{0.27^3} times higher which is equal to 50.81.

It means  in a cube that today contains one galaxy the size of the Milky Way, we would instead find 50.81 galaxies this size.

You can round this off to 52

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An object appears red when viewed with white light. What color would it appear when viewed with blue light?
zvonat [6]

Answer:

Colour of any object is by the colour of light it reflects i.e if white light is incident on the object,it will reflect blue color.so it will appear blue.But if red light is incident on it,it will not reflect that and absorb it.so as it will not reflect any light it will appear black.

Explanation:

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3 years ago
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What does the law of conservation of energy state?
pishuonlain [190]
The answer is B. good luck :)
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shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; r
Ainat [17]

Answer:

\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}

Explanation:

Given that

Charge on ring 1 is q1 and radius is R.

Charge on ring 2 is q2 and radius is R.

Distance ,d= 3 R

So the total electric field at point P is given as follows

Given that distance from ring 1 is R

\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0

\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(3R-R)}{(R^2+4R^2)^{3/2}}=0

\dfrac{q_1R}{(2R^2)^{3/2}}-\dfrac{q_2(2R)}{(5R^2)^{3/2}}=0

\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}

8 0
3 years ago
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