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kherson [118]
3 years ago
7

Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-3.4 μC). Charge q2 is located at

the origin and charge q6 is located a distance d = d1 + d2 = 7.4 cm from the origin as shown. What is ΔPE, the change in potential energy now if charge q1 is moved from point P to point R? d2=2.1cm d1=5.3cm, q1=1.5microC
Physics
1 answer:
ivanzaharov [21]3 years ago
3 0
I believe the answer would be zero because the q1 and q2 are equal on opposite sides and it

hope this helps
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The equilibrium temperature of the surface of the moon can be found by the formula as follows:

T=(\frac{K_s(1-alebdo)}{4 \sigma})^{\frac{1}{4}}

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<u>Explanation:</u>

Given-

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After 1 minute, thermometer reads 140°F. So,

140 = Tm + (65 - Tm)e^k                     - 2

Dividing equation 2 by 1:

e^k^-^0^.^5^k = \frac{140 - Tm}{110 - Tm} \\\\e^0^.^5^k = \frac{140 - Tm}{110 - Tm}                                - 3

From 1 we have,

e^0^.^5^k = \frac{110 - Tm}{65 - Tm}

Putting this vale in eqn 3. We get,

\frac{110 - Tm}{65 - Tm} = \frac{140 - Tm}{110 - Tm}

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