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umka2103 [35]
2 years ago
5

When an atom is involved in a nuclear reaction:

Physics
1 answer:
nataly862011 [7]2 years ago
4 0
I believe the answer would be , it may change from one element to another. I’m saying this because during nuclear transmutation is when a subatomic particle fired at the nucleus of an atom changes into a heavier element , to break the nucleus apart into two nuclei and energy.
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Which scientific term is defined as the rate at which work is accomplished?
Oliga [24]

Power is the scientific term defined as the rate at which work is accomplished. Power can be seen everywhere as it is applied to our everyday living. Power is seen when one pushes a heavy cart. It could be seen when a baseball hitter hits the ball.

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7 0
3 years ago
Plz guys answer my question
Sophie [7]

Answer:

Explanation:

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6 0
2 years ago
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In which one or more of the following is the earth assumed to be the center of the universe?
sladkih [1.3K]
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2 years ago
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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
1) La longitud del brazo de potencia de una palanca es de 0,0035 hectómetros y la del brazo de resistencia es de 55 centímetros.
Arlecino [84]

Answer:

Entonces seria 127 para vencer.

Explanation:

espero averte ayudado:-)

6 0
2 years ago
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