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3 years ago

5

Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a r

esting point midway between the two particles, an external agent must do work equal to

1 answer:

garri49 [273]3 years ago

8 0

Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J

Explanation:

Given that;

Mass M1 = 7.0 kg

r = 3.0/2 m = 1.5 m

Mass M2 = 21 kg

we know that G = 6.67 × 10⁻¹¹ N.m²/kg²

work done by an external agent W = -2GM2M1 / r

so we substitute

W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5

W = -1.96098 × 10⁻⁸ / 1.5

W = -1.3 × 10⁻⁸ J

Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J

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HELP ASAP HELP Two tennis balls of the same mass are served at different speeds: 30 m/s and 60 m/s. Which serve has more kinetic

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Answer:

<em>The second ball has four times as much kinetic energy as the first ball.</em>

Explanation:

<u>Kinetic Energy </u>

Is the type of energy an object has due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

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The kinetic energy of the first ball is:

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The kinetic energy of the second ball is:

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Being m the same for both balls, the second ball has more kinetic energy than the first ball.

To find out how much, we find the ratio:

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3 years ago

Read 2 more answers

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Answer:

Explanation:

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F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

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F2 = (F1 × r1^2)/(0.5 × r1)^2

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3 years ago

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Answer:

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Explanation:

From the question we are told that

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Generally the Mars atmosphere density is mathematically represented as

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Here A is the frontal area

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=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

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