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padilas [110]
3 years ago
10

Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate. what is the magnitude of the electric fi

eld e between the plates? express e in terms of q and other quantities given in the introduction, in addition to ϵ0 and any other constants needed.
Physics
1 answer:
jenyasd209 [6]3 years ago
5 0
<span>B?) Electric field E = q / (εo*A)

C) V = E*d = q*d / (εo*A)

D) V = q / C
so C = q / V = q * 1/V = q*εo*A / q*d = εo*A / d

I hope this helps! ^_^
</span>
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a mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test
choli [55]

F = 52000 N

m = 1060 kg

a= F/m = 52000 N/1060 kg = 49.0566 m/s^2

6 0
3 years ago
An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1000 K and a heat rejection at 400 K. During t
vivado [14]

Answer:

W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600

Explanation:

The Carnot cycle is described by

      e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}

     

In this case they indicate that the final volume is

         V = 3V₀

In the part of the heat absorption cycle from the source is an isothermal expansion

         W = n RT ln (V₀ / V)

         W / n = 8.314 1000 ln (1/3)

          W / n = - 9133 J / mol

During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times

            W / n = 8.314 400 (3)

           W / n = 3653 J / mol

The efficiency of the cycle is

            e = 1- 400/1000

            e = 0.600

6 0
3 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of
ratelena [41]

Answer:

Power output = 96.506 watts

Explanation:

Drag coefficient (Cd) = 0.9

V = 7.3 m/s

Air density (ρ) = 1.225 kg/m^(3)

Area (A) = 0.45 m^2

Let's find the drag force ;

Fd=(1/2)(Cd)(ρ)(A)(v^(2))

So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N

Drag power = Drag Force x Drag velocity.

Thus drag power, = 13.22 x 7.3 = 96.506 watts

8 0
3 years ago
When your body is warmed by an electric blanket during the winter, this process is said to be
lana66690 [7]
The correct answer would be A. endothermic because endothermic is-<span>(of a reaction or process) accompanied by or requiring the absorption of heat. Hope this helps!:) If you need any more just tag me.

</span>
8 0
3 years ago
Read 2 more answers
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