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Vsevolod [243]
3 years ago
6

Two particles collide, one of which was initially moving and the other initially at rest. Is it possible for both particles to b

e at rest after the collision?
Physics
1 answer:
solmaris [256]3 years ago
4 0

Answer:

Not possible

Explanation:

Unless there's some extra external force to keep both particles at rest after the collision, the momentum must be conserved before and after the collision.

So before the collision, 1 particle is at rest, 1 not ->  total momentum is non-zero

After the collision, both particles are at rest -> total momentum is zero which is different from before.

Therefore this is not possible.

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3 years ago
Which of the following would illustrate a quadratic relation between the dependent and independent variables when graphed?
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Answer: option A. a graph of the area of a circle vs. its radius r (A = πr²).



Explanation:



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This is it is a polynomial with general form ax² + bx + c, whewre a, b, and c, named coeficients,  are constants.



The function is y =  ax² + bx + c, where x is the independent variable and y is the dependent variable.



As stated in the question, the area of a circle is given by A = πr².



In this case, A is the dependent variable and r is the independent variable.



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The equation a = 1/b  is an inverse relation, not a quadratic relation.



The relation of distance vs. time for a car moving at constant speed is a linear relation of the kind v = u + st.



The mass of water vs. the volume of water in a drinking glass is a direct relation, mass = density × volume



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3 years ago
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4 years ago
A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a
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3 years ago
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The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso
nalin [4]

Explanation:

It s given that,

Mass of a planet, M=3.7\times 10^{24}\ kg

Radius of a planet, R=9.2\times 10^{6}\ m

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

g=\dfrac{GM}{R^2}

g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}

g=2.91\ m/s^2

(2) The escape velocity is given by :

v=\sqrt{\dfrac{2GM}{R}}

v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}

v = 7324.61 m/s

Hence, this is the required solution.

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