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3 years ago

11

An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

form electric field. What is the magnitude of the electric field if the field points 31.6° above the horizontal?

1 answer:

Tresset [83]3 years ago

4 0

Answer:231.16 N/C

Explanation:

Given

Electric Flux $=147 N-m^2/C$

Area(A) $=0.824 m^2$

Given Field point above $31.6 ^{\circ}$

Therefore angle between Area vector Electric Field =90-31.6= $58.4^{\circ}$

We know that Flux is given by

$\phi =\vec{E}\cdot \vec{A}$

$\phi =EAcos\theta$

$147=E\times 0.824\times cos(58.4)$

E=231.16 N/C

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1. A 0.40 kg ball is launched at a speed of 16 m/s from a 22 m cliff.

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Answer:

51.2 J, 86.2 J, 137.4 J

Explanation:

The kinetic energy of the ball is given by:

$K=\frac{1}{2}mv^2$

where

m = 0.40 kg is its mass

v = 16 m/s is its speed

Substituting,

$K=\frac{1}{2}(0.40)(16)^2=51.2 J$

The potential energy of the ball is given by

$U=mgh$

where

m = 0.40 kg

$g=9.8 m/s^2$ is the acceleration of gravity

h = 22 m is the heigth of the cliff

Substituting,

$U=(0.40)(9.8)(22)=86.2 J$

Finally, the total mechanical energy is the sum of the kinetic energy and the potential energy:

$E=K+U=51.2 + 86.2=137.4 J$

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3 years ago

Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard

podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

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Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

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