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lidiya [134]
2 years ago
8

Analyze the explanation and identify any false statements. Each statement to be analyzed is identified by a capital letter. A ba

l is rotating in a horizontal plane on the end of a string of length I. (A) The rotational kinetic energy remains constant as long as the length and angular speed are fixed. (B) When the ball is pulled inward and the length of the string is shortened, the rotational kinetic energy will remain constant due to conservation of energy, (C) but the angular momentum L will not remain constant because there is an external force acting on the ball to pull it inward. (D) The moment of inertia I remains constant (E) the angular speed will remain the same throughout the process because the ball is rotating in the same plane throughout the motion.
Physics
1 answer:
Sladkaya [172]2 years ago
4 0

Answer:

Explanation:

The rotational kinetic energy remains constant as long as the length and angular speed are fixed.

Statement A is true.

When the ball is pulled inward and the length of the string is shortened, the rotational kinetic energy will remain constant due to conservation of energy,

Statement B is false .

Reason - Conservation of energy will not be there because external work is done on the system  by the force that pulls it inward.  

but the angular momentum L will not remain constant because there is an external force acting on the ball to pull it inward

Statement C is false .

Reason - the angular momentum L will remain constant because there is an external force acting on the ball which acts perpendicular to the velocity of the ball .

The moment of inertia I remains constant

Statement D is false

Reason - because distance from axis of rotation is changing.

the angular speed will remain the same throughout the process because the ball is rotating in the same plane throughout the motion.

Statement E is false

Reason - Since moment of inertia decreases , to conserve angular momentum , angular speed increases.

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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

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                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

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3 years ago
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3 years ago
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SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}

Factoring out 2/3:

\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}

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7 0
3 years ago
Why is the spectrum of a star called an absorption spectrum?
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3 years ago
A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points
Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

(c) B-A = 15i -( -5j) = 15i + 5j

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

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