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Olenka [21]
3 years ago
15

Do you think it is easier or harder to hammer a nail into a floorboard on Pluto than on Earth? How about the Sun? Explain.

Physics
1 answer:
Serggg [28]3 years ago
7 0

Explanation:

It would be easier to lift the  hammer, during the nailing, in Pluto than on earth but it would require more hammer jacks to drive the nail through the  floorboard. This is because the gravity on Pluto is weaker than on Earth. The additional acceleration of the hammer (due to gravity) would be lesser hence you would require to put in a bit more energy for the downforce of the hammer.

On the sun, it would be difficult to lift the hammer than here on earth. This is because the gravity of the sun is much greater than on earth. If you would manage to lift the hammer, the downforce of the hammer on the nail would have an added acceleration of the gravity of the sun hence the force of the hammer hitting the nail would be higher hence rapidly driving it through the floorboard.

Learn More:

For more on gravity check out;

brainly.com/question/9934704

brainly.com/question/3034702

#LearnWithBrainly

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For any medium, other than vacuum, the index of refraction for red light is slightly lower (closer to 1 ) than that for blue light. This means that when light goes from vacuum (or air) into glass, the red light deviates from its original direction less than does the blue light. Also, as the light reemerges from the glass into vacuum (or air), the red light again deviates less than the blue light. If the two surfaces of the glass are parallel to each other, the red and blue rays will emerge traveling parallel to each other, but displaced laterally from one another.

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2 years ago
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A wire that has a resistance of 1Ω is to be built from an aluminum block of volume 1 cm3
alina1380 [7]

Answer:

  L = 5,955 m

Explanation:

For this exercise we must use the relation

        R = ρ L / A

where R is the resistance that indicates that it is 1 Ω, the resistivity is taken from the tables ρ = 2.82 10⁻⁸ Ω m, L is the length of the wire and A is the cross section.

As it indicates to us in volume of aluminum to use we divide the two terms by the length

    R / L = ρ L / (A L)

the volume of a body is its area times its length, therefore

  R / L = ρ L / V

  R = ρ L² / V

we clear the length of the wire

   L = √ R V /ρ

we reduce the volume to SI units

   v = 1 cm³ (1m / 10² cm)³ = 1 10⁻⁶ m

let's calculate

   L = √ (1  1  10⁻⁶ / 2.82 10⁻⁸)

   L = √ (0.3546 10²)

   L = 5,955 m

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