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allochka39001 [22]
3 years ago
12

Two capacitors have the same size of plates and the same distance 4 mm between the plates The potentials of the two plates in ca

pacitor 1 are 16 volts and 16 volts The potentials of the two plates in capacitor 2 are 320 volts and 352 volts a What is the electric field inside capacitor 1
Physics
1 answer:
OverLord2011 [107]3 years ago
8 0

Answer:

    E = -4000 N / C

Explanation:

The potential and electric field are related

         V = - E s

          E = - V / s

we reduce the magnitudes to the SI system

          s = 4 mm (1 m / 1000 mm) = 0.004 m

we calculate

          E = - 16 /0.004

          E = -4000 N / C

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Two particles each have the same mass but particle #1 has four times the charge of particle #2. Particle #1 is accelerated from
marin [14]

Answer:

 v_2 = 2*v  

Explanation:

Given:

- Mass of both charges = m

- Charge 1 = Q_1

- Speed of particle 1 = v

- Charge 2 = 4*Q_1

- Potential difference p.d = 10 V

Find:

What speed does particle #2 attain?

Solution:

- The force on a charged particle in an electric field is given by:

                                       F = Q*V / r

Where, r is the distance from one end to another.

- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:

                                      F_net = m*a

- Equate the two expressions:

                                      a = Q*V / m*r

- The speed of the particle in an electric field is given by third kinetic equation of motion.

                                      v_f^2 - v_i^2 = 2*a*r

Where, v_f is the final velocity,

            v_i is the initial velocity = 0

                                      v_f^2 - 0 = 2*a*r

Substitute the expression for acceleration in equation of motion:

                                       v_f^2 = 2*(Q*V / m*r)*r

                                       v_f^2 = 2*Q*V / m

                                       v_f = sqrt (2*Q*V / m)

- The velocity of first particle is v:

                                       v = sqrt (20*Q / m)

- The velocity of second particle Q = 4Q

                                       v_2 = sqrt (20*4*Q / m)

                                       v_2 = 2*sqrt (20*Q / m)

                                       v_2 = 2*v  

3 0
3 years ago
What information is given by the formula of an ionic compound?
Sergio [31]
MgCl2 is ionic compound.........Mg +2 and Cl -1
both charges are cross multiplied to each element......formula tells us that to balance the positive and negative charges on both sides they are cross multiplied........MgCl2......meaning there is one atom of Mg and 2 atoms of Cl.......

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4 0
3 years ago
Two people are trying to pull a 5-kilogram block in opposite directions. The first person pulls to the right with a force of 25
klasskru [66]

TEST TEST TEST TEST TEST TEST TEST

7 0
3 years ago
In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75kN. Knowing that v = 0.30 and E =
VladimirAG [237]

Answer:

(a) Elongation of rod = 0.19732 mm

(b) Change in diameter = 0.00651 mm

Explanation:

Circular area at end of steel rod = pi * diameter^2 / 4

Area = pi * (22 * 10^-^3)/4

Area = 3.801 * 10^(-4)    meter squared

Stress = force / area

Stress = 75000 / (3.801 * 10^-4)

Stress = 197316495 Pa     OR       0.197 GPa

Modulus of elasticity = stress / strain

200 = 0.19732 / Strain

Strain = 0.0009866     (longitudinal)

(a) Strain = change in length / initial length

0.0009866 = Elongation of rod / 200

Elongation of rod = 0.19732 mm

(b) Poisson ratio = lateral strain / longitudinal strain

0.3 = lateral strain / 0.0009866

Lateral strain = 0.000296

Lateral strain = Change in diameter / original diameter

0.000296 = Change in diameter / 22

Change in diameter = 0.00651 mm

4 0
3 years ago
Read 2 more answers
Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0
3 years ago
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