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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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The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision

The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, \Sigma P_{initial} is given as follows;

\Sigma P_{initial} = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ \Sigma P_{initial} = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ \Sigma P_{initial} = -20·i + 130·j

By the law of conservation of linear momentum, we have;

\Sigma P_{initial} = \Sigma P _{final} = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

\Sigma P _{final} = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

The magnitude of the final velocity, \left | v \right | = √((-0.08)² + (0.52)²) ≈ 0.526

\left | v \right | ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

4 0
3 years ago
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