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Svet_ta [14]
3 years ago
8

What happens when a charged insulator is placed near an uncharged metallic object?

Physics
2 answers:
AysviL [449]3 years ago
8 0
They attract each other. 
<span> When the charged insulator (regardless of whether + or -) is brought near an uncharged metallic object, electrons will migrate on the metallic object so that the area near the charged insulator is the opposite charge (and, by conservation of charge, the area of the conducting object away from the insulator will have an excess of the same charge as the insulator). 
</span><span> Because opposite charges attract and the opposite charge on the conducting object is closer, the dominant force will be one of attraction. 
</span><span> If they are allowed to touch, the conducting object will then have the same charge as the insulator and the objects will repel.</span>
Bad White [126]3 years ago
5 0
<span>So when the charged insulator is placed near the uncharged metallic thing they both attract each other because there will be a distribution of charge between the insulator and uncharged object. This migration of charges happens regardless of their charge positive or negative.</span>
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C.
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6 0
3 years ago
A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m bef
Dimas [21]

Supposing there's no air resistance, horizontal velocity is constant, which makes it very easy to solve for the amount of time that the rock was in the air.


Initial horizontal velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s 

15.5m / 10.3923m/s = 1.49s 

So the rock was in the air for 1.49 seconds. </span>

<span>

Now that we know that, we can use the following kinematics equation: 

d = v i * t + 1/2 * a * t^2 

Where d is the difference in y position, t is the time that the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>

<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s 

So: 

d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2 
d = 8.94 + -10.89</span>

d = -1.95<span>

<span>This means that the initial y position is 1.95 m higher than where the rock lands. </span></span>

5 0
3 years ago
Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th
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When they meet the 40-kg boy would have moved a distance of 6 m.

<h3>Distance moved by the 40 kg boy</h3>

Apply the principle of center mass;

Take the 40 kg mass as the reference point;

X(40 kg) = (40kg x 0  + 60kg x 10 m)/(40 kg + 60 kg)

X(40 kg) = (600)/(100)

X(40 kg) = 6 m

Thus, when they meet the 40-kg boy would have moved a distance of 6 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

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2 years ago
During a test crash and airbag in place to stop at dummies forward motion the dummies mass is 75 kg if the net force of the dumm
strojnjashka [21]
The dummy's acceleration is 11 m/s^2
(also known as 11 meters per sec. per sec.)

a = F/m
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3 0
3 years ago
Pls can someone help with question 37 asap pls
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The answer should be 2times3.14 which equals 6.28
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