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Maurinko [17]
3 years ago
5

Imagine that Kevin can instantly transport himself between Planet X and Planet Y. Which statement could be said about Kevin in t

his situation?
Physics
2 answers:
Over [174]3 years ago
8 0
What are the choices ? 

Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation.  A few of them might be . . .

-- Kevin will have no trouble getting back in time for dinner.

-- Kevin will have no time to enjoy the scenery along the way.

-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.

           -- Speed = (distance covered) / (time to cover the distance) .

If time to cover the distance is zero, then speed is huge (infinite).

           -- Kinetic energy = (1/2) (mass) (speed)² .

If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.

         -- Mass = (Kevin's rest-mass) / √(1 - v²/c²)

-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
-- If he succeeds somehow, his mass becomes imaginary.
-- At that point, he might as well turn around and go home ...
     if he ever reached Planet-Y, nobody could see him anyway.
vovangra [49]3 years ago
3 0

Kevin’s weight may change between Planet X and Planet Y.

Mass is the amount of matter contained in an object. Gravity is Earth’s pull of a body towards its center.  

Weight (W) is product of the mass (m) of the object and the force of gravity (g) acting on the object.  

W = mg  

The force of gravity may change between Plant X and Planet Y, and thus Kevin’s weight may change between Planet X and Planet Y.  

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If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by
lina2011 [118]

Answer:

The value is  v = 1.309*10^{5}\ m/s

Explanation:

The radius is r = 7.40 *10^{20} \  m

The mass of the ordinary matter is M_{rod} =  1.90 *10^{41}\  kg

Generally the speed of the star is mathematically represented as

         v = \sqrt{\frac{G * M}{r} }

Here G is the gravitational constant with a value

        G = 6.67384 * 10^{-11}

So

      v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }

=>    v = 1.309*10^{5}\ m/s

8 0
3 years ago
an athlete whirls a 7.00 kg hammer tied to the end of a 1.3 m chain in a horizontal circle the hammer moves at the rate of 1.0 r
ElenaW [278]

Answer:

The answer to your question is a = 1.3 m/s²

Explanation:

Centripetal acceleration is the motion of a body that transverse a circular path.

Data

mass = 7 kg

radius = r = 1.3 m

angular rate = w = 1.0 rev/s

centripetal acceleration = a = ?

Formula

a = rw²

Substitution

a = (1.3)(1)²

Simplification and result

a = 1.3 m/s²

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