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Serjik [45]
3 years ago
14

Please help me!!! ASAP 10 points for first to answer.

Physics
1 answer:
strojnjashka [21]3 years ago
7 0
See the image above :)

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3 years ago
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Four solid plastic cylinders all have radius 2.41 cm and length 5.94 cm. Find the charge of each cylinder given the following ad
Paladinen [302]

Answer:

Check explanation

Explanation:

QUICK NOTE: THE QUESTION IS NOT COMPLETE. Although it is not, we can make assumptions, since we only need values for the UNIFORM CHARGE DENSITY.

SO, LET US BEGIN;

To solve this question we are to use the equation (1) below;

Charge,Q = uniform charge density,p × Total area of the cylinder,A ------------------------------------------------------------------------(1).

From the question, we are given radius, R to be 2.41 cm and length, L to be 5.94 cm.

Step one: calculate for the total area of the cylinder, A.

Total area of the cylinder, A= area of the top surface + area of the buttom + area of the curved surface of the cylinder.

Hence, total area of the cylinder,A is;

==> πR^2 + πR^2 + 2πRL. -------------------------------------------------------------------------(2).

Then, total area of the cylinder,A is;

==> (L + R)2πR.

Step two: find the charge of each cylinder.

===> For the first cylinder; we have the uniform charge density to be 35 nC/m^2.

Therefore, the combination of equation (1) and (3) gives;

Charge Q= p × (L + R)2πR...----------------------------(4)

Hence, Q= 35 × [(5.94 + 2.41) 2× 3.143 × 5.94].= 10912.615 coulumb.

====> For the second cylinder, we have a uniform charge density of 50 nC/m^2.

Using equation (4), charge,Q= 15,589.45 Coulumb

=====> For THE third cylinder, the uniform charge density is 600, we make use of equation (4);

Charge,Q= 600×311.789.

Charge,Q= 187,073.4 coulumb.

====> For THE fourth cylinder, the uniform charge density is 750 nC/m^2.., we make use of equation (4);

Charge,Q= 233,841.75 coulumb.

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3 years ago
What happened to an iron bar if it is heated from 32 degree Celsius to 64 degree Celsius
olganol [36]

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it got hotter

Explanation:

it just did

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2 years ago
A circuit contains a single 270-pf capacitor hooked across a battery. it is desired to store four times as much energy in a comb
yan [13]
The energy stored by a system of capacitors is given by
U= \frac{1}{2}C_{eq} V^2
where Ceq is the equivalent capacitance of the system, and V is the voltage applied.

In the formula, we can see there is a direct proportionality between U and C. This means that if we want to increase the energy stored by 4 times, we have to increase C by 4 times, if we keep the same voltage.

Calling C_1 = 270 pF the capacitance of the original capacitor, we can solve the problem by asking that, adding a new capacitor with C_x, the new equivalent capacitance of the system C_{eq} must be equal to 4C_1. If we add the new capacitance X in parallel, the equivalent capacitance of the new system is the sum of the two capacitance
C_{eq} = C_1 + C_x
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C_1+C_x = 4C_1
from which we find
C_x=3C_1=3 \cdot 270 pF=810 pF
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