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nadezda [96]
3 years ago
6

Check My Answers Please!

Physics
1 answer:
GaryK [48]3 years ago
3 0
Good job correct you are smart 
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It has been suggested that rotating cylinders about 9 mi long and 5.9 mi in diameter be placed in space and used as colonies. Wh
mezya [45]

Answer:

\omega = 4.5\times 10^{-2} rad/s

Explanation:

Given data:

Rotating cylinder length = 9 mi

diameter of cylinder is 5.9 mi

we know that linear acceleration is given as

a =  r ω^2

where ω is angular velocity

so\omega = \sqrt{\frac{a}{r}}

r = \frac{5.9}[2} \frac{1609 m}{1 mi} = 4.746\times 10^{3} m

\omega = \sqrt{\frac{9.80}{4.746\times 10^{3}}}

\omega = 4.5\times 10^{-2} rad/s

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3 years ago
Galileo was a contemporary of
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Brahe & Kepler

Answer from Quizlet
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3 years ago
A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in
kykrilka [37]

Answer:

The child's mass is 14.133 kg

Explanation:

From the principle of conservation of linear momentum, we have;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂)  × v₃ - m₃ × v₄

We include the negative sign as the velocities were given as moving in the opposite directions

Since the child and the ball are at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Hence;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂)× v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

∴ m₁ + 2.4 = 16.533

m₁ = 16.533 - 2.4 = 14.133 kg

The child's mass = 14.133 kg.

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