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DiKsa [7]
3 years ago
13

If you have charged an electroscope by contact with a positively charged object, describe how you could use it to determine the

charge of other objects. Specifically, what would the leaves of the electroscope do if other charged objects were brought near its knob?
Physics
1 answer:
maxonik [38]3 years ago
7 0

Explanation:

  • When we have a pre-charged electroscope that is charged using a positive object then we observe that the leaves of the electroscope are spread wide in the diverging position due to the like charges of positive nature on the two leaves.
  • As we know that the charges can only travel through the conductors hence the assembly of the electroscope from the contact knob to the leaves is made of conductor. So the charges spread uniformly on this conductor.
  • Now we can use this as a tester of the other charges. When we bring a negatively charged object near the knob of the electroscope then the opposite charges i.e. positive charges accumulate near the knob due to attraction and the leaves of the electroscope go unstrained and get closer naturally.
  • Contrarily when a positive charge is brought near the knob of the electroscope then the the uniformly distributed charges move away from the knob and accumulate at the farthest point and more deviation is observed on the leaves.
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Alfredo makes a diagram to organize his notes about charging objects.
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The region marked X in the diagram shows that the objects have the same charge.

<h3>What is conduction?</h3>

The term conduction has to do with the manner of charging in which charge is passed from one object to another. Induction involves charging objects without the objects touching each other.

The region marked X in the diagram shows that the objects have the same charge.

Learn ore about conduction:brainly.com/question/15306642

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Which type of heat transfer causes your face to feel warm when you sit in the sun?
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A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the
ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

a_{B} =a_{D}  +(a_{B/D} )_{t}

2.5j=1.5j  +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

2\alpha =1\\\alpha =0.5rad/s^{2}CW

b) The acceleration of point A is:

a_{A} =a_{D}  +(a_{A/D} )_{t}

(aA/D)t = ADαj

a_{A} =a_{D}  +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}

The acceleration of point E is:

(aE/D)t = -EDαj

a_{E} =a_{D}  -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}

7 0
3 years ago
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