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Sergio039 [100]
4 years ago
6

The temperature -273°C is —.

Physics
1 answer:
ArbitrLikvidat [17]4 years ago
4 0

Answer:

-273°C = Absolute zero

Explanation:

Average Kinetic energy of particles in matter refers to the temperature of matter. This value is not arbitrary zero temperature. It explains the lowest kinetic energy of the particles in matter.

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Read 2 more answers
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
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