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scZoUnD [109]
3 years ago
11

Guys they deleted my account i was charliecole7 i had to make a new one

Physics
2 answers:
xxTIMURxx [149]3 years ago
4 0

Answer: ok???

Explanation:

Maslowich3 years ago
4 0
We didnt ask lololololol
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What is the orbital period in years of a planet with a semi major axis of 35 au
mezya [45]

Answer:

Orbital period of the planet will be 207.06 year                      

Explanation:

We have given the planet have the semi major axis as 35 au

We have to find the orbital period of the planet

From Keplar's third  law there is relation between the orbital period and semi major axis which is t T^2=R^3

So T^2=35^3

T^2=42875

T=207.06year

So orbital period of the planet will be 207.06 year

6 0
3 years ago
A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located
zhenek [66]

The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁  is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241  \times 10^{-19 } \ C

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

6 0
2 years ago
What times what equals 400
Radda [10]

If you're willing to consider fractions or decimals,
then there are an infinite number of answers. 
Like (2.5 x 160), and (15 x 26-2/3).

If you want to stick to only whole numbers,
then these 8 combinations do:

1, 400
2, 200
4, 100
5, 80
8, 50
10, 40
16, 25
20, 20

7 0
4 years ago
What is an example of an average main squence star?
Aleks [24]
Main sequence stars fuse hydrogen atoms to form helium atoms in their cores. About 90 percent of the stars in the universe, including the sun, are main sequence stars. These stars can range from about a tenth of the mass of the sun to up to 200 times as massive. Stars start their lives as clouds of dust and gas.
5 0
3 years ago
Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec
Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is \dfrac{11}{17}\ J

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value intro the formula

300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v

v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}

v=-0.5\ m/s

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2

K.E_{i}=0.31875\ J

We need to calculate the total final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2

K.E_{f}=0.1125\ J

We need to calculate the energy lost during the collision

Using formula of energy lost

energy\ lost=\dfrac{0.31875-0.1125}{0.31875}

energy\ lost=\dfrac{11}{17}\ J

Hence, The fraction of the total initial kinetic energy is lost during the collision is \dfrac{11}{17}\ J

3 0
3 years ago
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