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Doss [256]
3 years ago
13

A uniform, thin, solid door has height 2.20 m, width 0.870 m, and mass 23.0 kg. (a) find its moment of inertia for rotation on i

ts hinges. (b) is any piece of data unnecessary
Physics
1 answer:
vampirchik [111]3 years ago
8 0
The solution for this problem is:


a. The moment of inertia of the thin door rotating about the hinges axis is given by:

I = 1/3 mw^2 is to solve for the inertia, where: m is the mass and w is the width


= 23.0 x 0.870^2 / 3

= 17.4087 / 3

= 5.8029 kg m^2


b. The height of the door is not required for this calculation.
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sergiy2304 [10]

Answer:

1. Golf ball is in the range of only particulate detectable range

2, alpha partcle is in ultraviolet range of wavelength

3.bullet is in xray range of wavelength

All in the EMW spectrum

6 0
3 years ago
A gecko crawls vertically up and down a wall. Its motion is shown on the following graph of vertical position yyy vs. time ttt.
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4 0
3 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
3 years ago
Calculate the AMA of an access ramp if it takes 255 N of force to push a person in a wheelchair having a combined weight of 764
Andreyy89
Actual Mechanical Advantage(AMA) = Weight / Force
Here, Weight = 764 N
Force = 255 N

Substitute the values in to the expression, 
AMA = 764 / 255
AMA = 2.99

After rounding-off to the nearest tenth value, it would be 3

Finally, option C would be your answer.

Hope this helps!
7 0
3 years ago
Two identical bodies are sliding toward each other on a frictionless surface. One moves at 1 m/s and the other at 2 m/s. They co
mario62 [17]

Answer:

V=\dfrac{3}{2}\ m/s

Explanation:

Two identical bodies are sliding toward each other on a frictionless surface.

Initial speed of body 1, m₁ = 1 m/s

Initial speed of body 2, m₂ = 2 m/s

They collide and stick.

We need to find the speed of the combined mass. Let V is the speed of the combined mass.

Using the conservation of momentum.

m_1u_1+m_2u_2=V(m_1+m_2)

We have, m₁ = m₂ = m

m\times 1+m\times 2=(m+m)V\\\\m+2m=2m\times V\\\\3m=2mV\\\\V=\dfrac{3}{2}\ m/s

So, the speed of the combined mass is \dfrac{3}{2}\ m/s.

6 0
2 years ago
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