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Studentka2010 [4]
3 years ago
10

An explosion that occurs at the end of a massive star's life is a(n)?

Physics
1 answer:
Ahat [919]3 years ago
3 0
<span>An explosion that occurs at the end of a massive star's life is a supernova</span>
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True or False?
lana66690 [7]
The answer is A. True 
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4 years ago
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Gravity is affected by the ______
Alla [95]

C is the right answer :)

C. mass of the objects and the distance between two objects

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. A newly discovered planet has three times the mass and five times the radius of Earth. What is the ratio of the acceleration d
NikAS [45]

Answer:

0.12

Explanation:

The acceleration due to gravity of a planet with mass M and radius R is given as:

g = (G*M) / R²

Where G is gravitational constant.

The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg

The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m

Therefore:

g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²

g(planet) = 1.18 m/s²

Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:

g(planet)/g(earth) = 1.18/9.8 = 0.12

3 0
3 years ago
When elements react they can form molecules. What else might they form?
MrMuchimi

Answer:

they react together and form compounds

Explanation:

3 0
3 years ago
Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x
GalinKa [24]

Answer:

a) (-367231.63i ,  367231.63i, 0) N/C

b) (0 , 0  , 367231.63i ) N/C

Explanation:

a)

Case x < -2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i

Case x > 2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

Case -2.15 < x <+2.15

E =E_{+} - E_{+} \\\\E = 0

b)

Case x < -2.15

E =E_{+} - E_{+} \\\\E = 0

Case x > 2.15

E =E_{+} - E_{+} \\\\E = 0

Case -2.15 < x <+2.15

E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

7 0
3 years ago
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