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3 years ago

A 3.0-kilogram cart possesses 96 joules of kinetic energy. Calculate the speed of the car.

Physics

1 answer:

Dvinal [7]3 years ago

5 0

Answer:

8.0 m/s

Explanation:

KE = ½ mv²

96 J = ½ (3.0 kg) v²

v = 8.0 m/s

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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

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