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yarga [219]
3 years ago
12

The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro

jectile is shot to a height of 5.0 m above the end of the expanded spring. (See below.) How much was the spring compressed initially?
Physics
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

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Answer:

The car's displacement during this time is 25.65 meters.

Explanation:

Given that,

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7 0
3 years ago
Sally and Sam are in a spaceship that comes to within 16,000 km of the asteroid Ceres. Determine the force Sally experiences, in
Diano4ka-milaya [45]

Answer:

  0.017 N

Explanation:

The relevant relation is ...

  F = GMm/r²

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Filling in the given numbers, we find the force to be ...

  F = (6.67408 × 10^-11 m^3·kg^-1·s^-2)(8.7 × 10^20 kg)(77 kg)/(1.6 × 10^7 m)^2

where m in this expression is the unit "meters".

  F = 6.67408 · 8.7 · 77/2.56 × 10^(-11 +20 -2·7) N ≈ 0.017 N

The asteroid exerts a force of about 0.017 N on Sally.

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<em>Additional comment</em>

That's about 0.000023 times the force of Earth's gravity.

7 0
2 years ago
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