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abruzzese [7]
3 years ago
6

You have landed on an unknown planet, Newtonia, and want to know what objects weigh there. When you push a certain tool, startin

g from rest, on a frictionless horizontal surface with a 12.0-N force, the tool moves 16.0 m in the first 2.00 s. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia, and what does it weigh on Earth?
Physics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

Explanation:

Given

Force applied F=12\ N

time taken t_1=2\ s

Displacement s=16\ m

using

s=ut+\frac{1}{2}at^2

u=initial velocity

s=displacement

t=time

16=0+\frac{1}{2}a(2)^2

a=8\ m/s^2

thus mass of body m=\frac{F}{a}=\frac{12}{8}=1.5\ kg

Next it is released from a height of h=10\ m

time taken to reach ground t_2=2.58\ s

using  s=ut+\frac{1}{2}at^2 in vertical direction

here acceleration is due to acceleration due to gravity(g') of the planet

as it at rest so u=0 here

10=0+\frac{1}{2}(g')(2.58)^2

g'=3.004\ m/s^2

Thus acceleration due to gravity on Newtonian is 3\ m/s^2

Weight of tool on Newtonia W'=mg'

W'=1.5\times 3.004=4.506\ N

Weight on Earth W=1.5\times 9.8=14.7\ N

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Likurg_2 [28]

Answer:

Approximately \rm 11.2 \; N \cdot kg^{-1} at that distance from the center of the planet.

Option A) The low value of g near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of g on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

\displaystyle \frac{G \cdot M \cdot m}{R^2},

where

  • G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2.
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  • m is the mass of the object.

To find an equation for g, divide the equation for gravity by the mass of the object:

\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}.

In this case,

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Calculate g based on these values:

\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}.

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for g:

\displaystyle g = \frac{G \cdot M}{R^2}.

The mass of the planet is in the numerator. If two planets are of the same size, g would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since R is in the denominator of g, increasing the value of R while keeping M constant would reduce the value of g. That explains why the value of g near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately 9.81\; \rm N \cdot kg^{-1}.

As a side note, 5.82\times 10^7\rm \; m likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of g near the cloud tops of Saturn is approximately \rm 11.2 \; N \cdot kg^{-1}.

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People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th
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Answer:

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I am pretty sure it is B....
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Answer:

65 m/s

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