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Dmitriy789 [7]
3 years ago
5

How does weight change as the gravitational acceleration changes and why?

Physics
1 answer:
seraphim [82]3 years ago
4 0

Answer

The correct formula for weight is F = m*g where g is the gravitational acceleration.

All over the earth's surface, g is slightly different. The mass does not change no matter where you are. We should be measuring out weights in Newtons, not in kg. So the unit of weight in the metric system is 9.8 about * mass in kg.

Stop reading. This is your answer.

============

Notes

It was hard enough to get people to change over to kg never mind newtons. Canada, which is on the metric system, still gives the price of food in pounds. Or if not, in grams if the container is small enough.

Cashews cost 19$ Canadian per 906 grams which is roughly 2 pounds.

Oranges are $1.27 a pound and that is the way they are listed in Walmart.

10 pounds of potatoes are 4.95 dollars.

I'm sure you get the point. We use kg for certain things and retain pounds for others.  

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A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o
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Answer:

Recoil speed, 1.17\times 10^{-5}\ m/s                          

Explanation:

Given that,

Mass of the comet fragment, m_1=1.96\times 10^{13}\ kg

Speed of the comet fragment, v_1=6.5\times 10^4\ m/s

Mass of Callisto, m_2=1.08\times 10^{23}\ kg

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

m_1v_1=(m_2+m_2)V

V is recoil speed of Callisto immediately after the collision.

V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s

So, the recoil speed of Callisto immediately after the collision is 1.17\times 10^{-5}\ m/s

6 0
3 years ago
Expansionary monetary policies would likely cause
Jlenok [28]

Answer:

TRUE

Explanation:i've seen one and please make me the brainlest

6 0
2 years ago
Suggest a situation where we can obtain more than one shadow of an object at a time?
Murljashka [212]

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We can obtain more than one shadow of an object if light from more than one source falls on it. [For example during a match being played in a stadium multiple shadows of players are seen].

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Notice that we took the start time to be the start of the 5 second interval and set that to t_0=0. The starting velocity v_0 is the velocity measured at the start of the interval, and v is the velocity measured at its end.

So the average velocity over these 5 seconds is

\bar a=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}{5\,\mathrm s}=6\,\dfrac{\mathrm m}{\mathrm s^2}

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faust18 [17]
A is the correct answer.

please vote my answer branliest! Thanks.
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4 years ago
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