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Airida [17]
2 years ago
11

In each case, a charged rod, made of the dense rubber ebonite, comes close or is in contact with the top of an electroscope. The

ball on top of the electroscope is directly connected to the two metal leaves suspended in the flask. Which image represents a gaining of a charge on the leaves of the electroscope by conduction In each case, a charged rod, made of the dense rubber ebonite, comes close or is in contact with the top of an electroscope. The ball on top of the electroscope is directly connected to the two metal leaves suspended in the flask. Which image represents a gaining of a charge on the leaves of the electroscope by conduction? A) A B) B C) C D) D
Physics
2 answers:
Art [367]2 years ago
8 0

Answer:

b

Explanation

B represents a charge gained by induction. In this case, the rod is not touching the electroscope and the leaves are charged similarly because they are repelling each other. The charge is gained (induced) by the nearness of the oppositely charged rod.

on:

Evgen [1.6K]2 years ago
5 0

Answer:

D

Explanation:

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b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

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How can the motion of an object that is NOT moving change?​
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Only if a force acts upon it, it can move.
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3 years ago
A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp
AlekseyPX

Answer:

(A) 2.4 N-m

(B) 0.035kgm^2

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque \tau =Fr, here F is force and r is distance

So \tau =20\times 0.12=2.4Nm

(B) Moment of inertia is equal to I=\frac{1}{2}mr^2

So I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2

Torque is equal to \tau =I\alpha

So angular acceleration \alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2

(C) As the disk starts from rest

So initial angular speed \omega _{0}=0rad/sec

Time t = 4.6 sec

From first equation of motion we know that \omega =\omega _0+\alpha t

So \omega =0+68.571\times  4.6=315.426rad/sec

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(E) From second equation of motion

\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad

3 0
3 years ago
you measured the torque caused by the gyroscope's weight by lifting up the end of the gyroscope (at 7.5 inches). If you measured
DIA [1.3K]

Answer:

i) No, the spring scale does not read a different value

ii) The torque will read a different value, it will reduce

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Explanation:

The torque caused by the gyroscope can be given by the relation,

\tau = r × f

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The torque measured by the gyroscope varies directly with the distance, r.

A decrease in the distance r will also cause a decrease in the value of the torque measured. When the distance, r is reduced from 7.5 inches to 5 inches, the torque caused by the gyroscope's weight also reduces.

The weight of the gyroscope remains constant despite the reduction in the distance because the weight  of the gyroscope is not a function of the distance from the gyroscope. Therefore, the spring scale will not read a different value.

Yes, the spring scale does not need to be measured from the center of mass location because the weight does not depend on the location of measurement. The reading of the sprig scale remains constant.

3 0
3 years ago
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