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stellarik [79]
3 years ago
7

If element X has 17 protons, how many electrons does it have?

Physics
1 answer:
OleMash [197]3 years ago
3 0
18 electrons because protons minus the atomic mass so it would be 35-17 which gives you 18.
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I need an answer asap
gulaghasi [49]
He has a mass of 56 kg.

The equation given is PE = mgh.

PE = 4620 J

h = 8.4

g = 9.8

Therefore:

4620 = 82.32m

m = 4620/82.32
m = 56 (rounded to two significant digits)
5 0
3 years ago
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12
Licemer1 [7]

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

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

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

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         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}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

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}  

         K = 20 10⁻⁹ J

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

3 0
2 years ago
Why is the distance between the sun and Neptune dim light years
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The average distance from the Sun to Neptune is about 2.795 billion miles.

That's roughly  0.00048  of a light year .

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