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3 years ago

Is it possible to do work on an object without changing the kinetic energy of the object? Now Why?

Physics

1 answer:

Murrr4er [49]3 years ago

8 0

Answer:

(a) Yes, it is possible by raising the object to a greater height without acceleration.

Explanation:

The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.

If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.

If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).

Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.

Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".

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A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

3 years ago

A certain organ pipe, open at both ends, produces a fundamental frequency of 288 Hz in air. Part A If the pipe is filled with he

alina1380 [7]

Answer:

773.25 Hz

Explanation:

Concept : In an open organ pipe in fundamental mode of vibration

wave length of wave λ = 2L

where L is length of the pipe

frequency = velocity of sound / λ

Given values: fundamental frequency = 288 Hz

fluid is air. velocity of sound = 340 m/s

⇒ 288 = 340/2L

⇒L = 59.02 cm

The point to be noted is if the pipe is filled with helium initially at the same temperature, there would be change in the sound velocity .Then, frequency of note produced will also be changed .

We know that velocity of sound is inversely proportional to square root of molar mass of gas

velocity of sound in air / velocity of sound in helium = Square root of (Molar mass of Helium/ molar mass of air)

$\frac{V_a}{V_{He}} = \sqrt{\frac{4}{28.8} } \\\frac{340}{V_{He}} =0.3725\\V_{He} =912.5 m/s$

Now, frequency = velocity of sound / λ

= 912.75 / (2 x 0.5902)

= 773.25 Hz

6 0

3 years ago

There are two particles with charges +Q and +q. The electric force applied to one of the charges is 10 N. Now, +Q is replaced by

KIM [24]

Answer:

F = 10 N

Explanation:

Assuming that we can treat to both particles as point charges, the magnitude of the force that one charge exerts on the other, must obey Coulomb's Law.
The expression for this force, applied to the charges +Q and +q, separated by a distance d, is as follows:

$F_{0} = \frac{k*q*Q}{d^{2}}$