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3 years ago

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An electric motor spins at a constant 2695.0 rpm. If the rotor radius is 7.165 cm, what is the linear acceleration of the edge o

f the rotor?A) 5707 m/s^2 B) 281.6 m/s^2 C) 28.20 m/s^2 D) 572,400 m/s^2

1 answer:

Vikki [24]3 years ago

5 0

Answer:

5707 m/s^2

Explanation:

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Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0×1011 N/m2. A cube of steel 25 cm on a side suppor

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Answer:

Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m

Explanation:

Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg

Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.

We know from Newton's Second Law that

F = ma where m is the mass and a is the acceleration (<em>in this case due to gravity</em>) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So

F (normal) = M (load) x a = 85 x 9.81 = 833.85 N

Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by

Y = (F/A)/(ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.

Solving this for ΔL, we can re- arrange the equation

ΔL = (F x L)/(Y x A) since area of square is L x L we can simplify the equation to get

ΔL = (F)/(Y x L), substitute the values

ΔL = (833.85)/(2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m

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3 years ago

Read 2 more answers

A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodie

vivado [14]

Answer:

The length of the rod for the condition on the question to be met is $L = 1.5077 \ m$

Explanation:

The Diagram for this question is gotten from the first uploaded image

From the question we are told that

The mass of the rod is $M = 3.41 \ kg$

The mass of each small bodies is $m = 0.249 \ kg$

The moment of inertia of the three-body system with respect to the described axis is $I = 0.929 \ kg \cdot m^2$

The length of the rod is L

Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

$I = I_r + 2 I_m$

Where $I_r$ is the moment of inertia of the rod about the describe axis which is mathematically represented as

$I_r = \frac{ML^2 }{12}$

And $I_m$ the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as

$I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}$

Thus $2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}$

Hence

$I = M * \frac{L^2}{12} + m * \frac{L^2}{2}$

=> $I = [\frac{M}{12} + \frac{m}{2}] L^2$

substituting vales we have

$0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2$

$L = \sqrt{\frac{0.929}{0.40867} }$

$L = 1.5077 \ m$

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