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3 years ago

Determine the distribution of charges on the sphere if it is made of non-conducting plastic.

Physics

1 answer:

Elden [556K]3 years ago

7 0

Answer:

If the sphere is made of a non-conducting material, charges can´t distribute freely over its surface or volume. In non-conducting material bodies, charges can be transported from a molecule to another nearest to it. Therefore if you rub an object with a positive charge with this sphere made of non-conducting plastic, the object will rip some negative charges (electrons) of that zone of the sphere. Therefore that zone will get a superficial positive charge density, but the non-affected zone of the sphere will keep having a null charge density.

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jasenka [17]

Speed of Ferry is towards North with magnitude 6.2 m/s

Here if we assume that North direction is along Y axis and East is along X axis then we can say

$\vec v_f = 6.2 \hat j$

Now a person walk on ferry with speed 1.5 m/s towards east with respect to Ferry

so it is given as

$\vec v_{pf} = 1.5 \hat i$

also by the concept of relative motion we know that

$\vec v_{pf} = \vec v_p - \vec v_f$

now plug in all values in it

$1.5 \hat i = \vec v_p - 6.2 \hat j$

$\vec v_p = 1.5 \hat i + 6.2 \hat j$

now if we need to find the speed of the person then we need to find its magnitude

so it is given as

$v = \sqrt{1.5^2 + 6.2^2}$

$v = 6.37 m/s$

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3 years ago

On Mars a rock falls an unknown vertical distance from a resting position and lands in a crater. If it takes the rock 2.5 second

astra-53 [7]

The Answer To This Question Is B

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3 years ago

S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

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$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

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3 years ago

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2 years ago

You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

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From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

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