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kykrilka [37]
3 years ago
12

How many days does it take for a free to grow?

Physics
1 answer:
vovikov84 [41]3 years ago
7 0
Idk what is growing but if it’s a free than c
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Which option is the best blackbody radiator?
SSSSS [86.1K]

Answer:

A. The Sun

Explanation:

The Sun is to be considered a perfect black body.

7 0
3 years ago
How to get infinite thanks could you show me please ☺️​
murzikaleks [220]

Answer: divide by zero, or square root of a negative

Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.

A couple of ways to get that...

  1. You try to divide by zero. In other words, a problem that asks you to perform something like this:  5/0=   or 23/(4-4)=   Such a problem will give you an error on a calculator because the answer is infinity or doesn't exist.
  2. Another way is to try to get the square root of a negative number. That answer doesn't exist as a real number, so \sqrt{-4\\} will also give you an error on a calculator.

7 0
3 years ago
g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

T^2 = \frac{4\pi^2 d^3}{2GM}

The given data are given with respect to known constants, for example the mass of the sun is

m_s = 1.989*10^{30}

The radius between the earth and the sun is given by

r = 149.6*10^9m

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

m = 8.2*1.989*10^{30}

d = 6.2*149.6*10^6

Substituting in Kepler's third law:

T^2 = \frac{4\pi^2 d^3}{2}

T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}

T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}

T = 120290789.7s

T = 120290789.7s(\frac{1year}{31536000s})

T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
3 years ago
Two charged particles, Q 1 and Q 2, are a distance r apart with Q 2 = 5 Q 1. Compare the forces they exert on one another when i
Sergio [31]
F2=-F1 !!!!!!!!!! Hope it helps
8 0
4 years ago
Earth reads 650N what is your mass
sesenic [268]
P = m . g
650 = m.10
m = 650/10
m = 65 kg
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3 years ago
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