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Svetach [21]
4 years ago
8

A uniform, thin, solid door has height 2.30 m, width 0.875 m, and mass 24.0 kg. (a) find its moment of inertia for rotation on i

ts hinges.
Physics
1 answer:
vekshin14 years ago
3 0
A uniform thin solid door has height 2.20 m, width .870 m, and mass 23.0 kg. Find its moment of inertia for rotation on its hinges. Is any piece of data unnecessary? So far, I don't understand how to calculate moments of inertia for things like this at all. I can do a system of particles, but when it comes to any ridgid objects, such as this door or rods or cylinders, I don't get it. So basically I have no idea where to even start with this. 

so A
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Complete the paragraph to compare Uranus and Neptune.
kipiarov [429]

Answer:

Uranus and Neptun are outer planets od the Solar system, since they are located after the asteroid belt. All of these outer planets are much larger then the inner ones so they are called the "ice giants". The other reason for this name is that they are very far from the Sun, so their temperature is low. Another feature they have in common is their atmosphere which is composed of gases, including methane, which is responsible for their blue color, since methane absorbs red light. However Neptune is known for very fast winds and storms in its atmosphere which is responsible for its high activity and changes.

So, the blanks should be filled with:

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4 years ago
How is energy and work related.<br>​
Anna007 [38]

Answer:

Varies

Explanation:

They both relate to the process of doing something.

5 0
3 years ago
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What is the resultant of the vectors shown?
adelina 88 [10]

the one with the big mustard-colored stripe across it is the correct choice.

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3 years ago
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A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is conne
kipiarov [429]

Answer:

0.832 m/s

Explanation:

The work done by the spring W equals the kinetic energy of the object K

The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀   = initial compression = 0.065 m and x₁ = final compression = 0.032 m

The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object

Since W = K,

1/2k(x₀² - x₁²) = 1/2mv²

k(x₀² - x₁²) = mv²

mv² = k(x₀² - x₁²)

v² = [(k/m)(x₀² - x₁²)]

taking square root of both sides, we have

v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),

v = √[(k/m)√(x₀² - x₁²)]

v = ω√(x₀² - x₁²)]

Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have

v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]

v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]

v = 14.7 rad/s × √(0.003201 m²)

v = 14.7 rad/s × 0.056577

v = 0.832 m/s

8 0
3 years ago
What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation?.
Ber [7]

Answer:

It has no effect on the amplitude.

Explanation:

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