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GenaCL600 [577]

1 year ago

9

find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sus

tainable electric field strength in air to be 3 x 106 v/m. .

1 answer:

Vlada [557]1 year ago

6 0

The maximum potential difference between two parallel conducting plates is 1.5×10^4 V.

As

Electric field strength is: E = 3×10^6 V/m

distance d = 5×10^-3 m

To find Potential Difference V,

E = V/d

or

V = Ed

Putting the values

V = 3×10^6 × 5×10^-3

V = 1.5×10^4 V

The maximum potential difference between two parallel conducting plates is 1.5×10^4 V.

If you need to learn more about Potential difference, click here

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NO LINKS. We awoke at the uncomfortably early hour of 5:30 a.m. Our eyes were attacked by the unforgiving rays of the rising Sun

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B. The writer didn't mention anything significant .

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3 years ago

A ball of mass m= 450.0 g traveling at a speed of 8.00 m/s impacts a vertical wall at an angle of θi =45.00 below the horizontal

Mkey [24]

Answer:

F = 20.4 i ^

Explanation:

This exercise can be solved using the ratio of momentum and amount of movement.

I = F t = Dp

Since force and amount of movement are vector quantities, each axis must be worked separately.

X axis

Let's look for speed

cos 45 = vₓ / v

vₓ = v cos 45

vₓ = 8 cos 45

vₓ = 5,657 m / s

We write the moment

Before the crash p₀ = m vₓ

After the shock $p_{f}$ = -m vₓ

The variation of the moment Δp = mvₓ - (-mvₓ) = 2 m vₓ

The impulse on the x axis Fₓ t = Δp

Fₓ = 2 m vₓ / t

Fx = 2 0.450 5.657 / 0.250

Fx = 20.4 N

We perform the same calculation on the y axis

sin 45 = vy / v

vy = v sin 45

vy = 8 sin 45

vy = 5,657 m / s

We calculate the initial momentum po = m $v_{y}$

Final moment $p_{f}$ = m $v_{y}$

Variations moment Δp = m $v_{y}$ - m $v_{y}$ = 0

Force in the Y-axis $F_{y}$ = 0

Therefore the total force is

F = fx i ^ + Fyj ^

F = Fx i ^

F = 20.4 i ^

3 0

4 years ago

The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge

Eva8 [605]

Answer:

a) $v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) $v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) $v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) $v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) $v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

$KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)$ (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

$\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}$

We can write the mass of a proton in MeV/c².

$m_{p}=938.28 MeV/c^{2}$

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.04$

$v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) Linac (400 MeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.71$

$v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) Booster (8 GeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.994$

$v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) Main ring or injector (150 Gev)

$\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.999$

$v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) Tevatron (1 TeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.9999$

$v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Have a nice day!

4 0

4 years ago

El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La

allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta$

En donde:

$\omega_{f}$ : es la velocidad angular final = 2000 rpm = 209,4 rad/s

$\omega_{0}$ : es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.

Resolviendo la ecuación (1) para α, tenemos:

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}$

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.

b) El tiempo transcurrido se puede encontrar como sigue:

$\omega_{f} = \omega_{0} + \alpha t$

Resolviendo para t, tenemos:

$t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s$

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

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3 years ago

Suppose two point charges, Q1 and Q2, are separated by a distance d. Which correctly states Coulomb's Law for the electrical for

likoan [24]

The correct answer is (B)

Which is (kQ1Q2) / d^2

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3 years ago