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k0ka [10]
3 years ago
12

An outside thermometer reads 57°F. What is this temperature in °C? Round your answer to the nearest whole number.

Physics
2 answers:
marishachu [46]3 years ago
3 0

Answer: It'd be 14.

Explanation:

The formula for this equation would be (57f-32)×5/9 which is equal to 13.889; and rounding that to the whole number would be 14.

Fiesta28 [93]3 years ago
3 0

Answer-

a) 14

Explanation:

Fahrenheit to Celsius Formula: (°F - 32) / 1.8 = °C

= [57-32]/1.8=C

=25/1.8=C

=13.8888888889

round off 13.8888888889=14

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The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

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        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

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         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

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The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

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In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

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Answer:

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Explanation:

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