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ANEK [815]
3 years ago
13

You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving

when it is halfway down the building? (assume no friction) (1 point)
20.78 m/s
39.81 m/s
19.68 m/s
26.92 m/s
Physics
1 answer:
BabaBlast [244]3 years ago
4 0

Y₀ = initial position of the balloon at the top of the building = 44 m

Y = final position of the balloon at halfway down the building = 44/2 = 22 m

a = acceleration of the balloon = - 9.8 m/s²

v₀ = initial velocity of the balloon = 0 m/s

v = final velocity of the balloon = ?

using the kinematics equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 0² + 2 (- 9.8) (22 - 44)

v = 20.78 m/s

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Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
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The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

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