What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

1 answer:

5 0

You could always directly use F = G M m / r², but that takes forever.

<span>It is best to know that, on Earth's surface, the astronaut would weigh </span>
<span>F = m g = 579.2 N </span>
<span>where m = 59.1 kg and g = 9.8 m/s². </span>

<span>You then notice from F = G M m / r² that doubling the mass of the earth would double his weight and doubling the radius would give 1/4 his original weight. So... </span>
<span>F = 1/4 * 2 * 579.2 N = 290 N </span>

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The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

3 years ago

That 1.5 kg brick falls to the ground. What is the kinetic energy when the brick is moving at 26 m/s? Formula: Variables: Work w

Tresset [83]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$