Ask question

Ask question

All categories

2 years ago

15

Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area

of radius 9.0 m that lies in the xy-plane? (Enter the magnitude.)

1 answer:

cluponka [151]2 years ago

4 0

Answer:

5.09 x 10⁵ Nm²/C

Explanation:

The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e

φ = E A

From the question;

E = (8.0j + 2.0k) ✕ 10³ N/C

r = radius of the circular area = 9.0m

A = area of a circle = π r² [Take π = 3.142]

A = 3.142 x 9² = 254.502m²

Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.

Therefore;

φ = (2.0) x 10³ x 254.502

φ = 5.09 x 10⁵ Nm²/C

The electric flux is 5.09 x 10⁵ Nm²/C

You might be interested in

What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an

astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution= $T_b=101^{\circ}$ C

Boiling point water=100 degree Celsius

$K_f=1.86K/m$

$K_b=0.512 K/m$

$\Delta T_b=T-T_0$

Where $T$ =Boiling point of solution

$T_0=$ Boiling point of pure solvent

$\Delta T_b=101-100=1^{\circ}$ C

$\Delta T_b=k_bm$

Using the formula

$1=0.512\times m$

Molality, $m=\frac{1}{0.512}$ m

$\Delta T_f=k_fm$

Using the formula

$\Delta T_f=\frac{1}{0.512}\times 1.86$

$\Delta T_f=3.63 C$

We know that

$\Delta T_f=T_0-T_1$

Where $T_0$ =Freezing point of solvent

$T_1=$ Freezing point of solution

Using the formula

$3.63=0-T_1$

Freezing point of water=0 degree Celsius

$T_1=0-3.63=-3.63 C$

Hence, the freezing point of solution=-3.63 degree Celsius

7 0

2 years ago

an 80 kg cart goes around the inside of a vertical loop of a roller coaster. the radius of the loop is 5m and the cart moves at

GarryVolchara [31]

Answer:

F= 224 N

Explanation:

Given that

mass ,m = 80 kg

Radius ,r= 5 m

speed at the top v= 8 m/s

The force at the top = F

Now by using the Second law of Newton's

$F+mg=\dfrac{mv^2}{r}$

Now by putting the values

$F+80\times 10 = \dfrac{80\times 8^}{5}$

Take g = 10 m/s²

$F= \dfrac{80\times 8^2}{5}-800$

F= 224 N

Therefore the force exerted by the track at the top position will be 224 N.

3 0

3 years ago

Fleas have remarkable jumping ability. a 0.60 mg flea, jumping straight up, would reach a height of 40 cm if there were no air r

irakobra [83]

To answer, we determine the velocity of the flea when it leaves the ground. Using the given value for distance, the velocity can be determined through the equation,

2a d = (Vf)² - (Vi)²

where a is the acceleration due to gravity, d is distance, Vf is the final velocity which is equal to zero and Vi is the initial velocity. Substituting the known values,

2(9.8 m/s²)(40cm/100cm) = (Vi)²

The value of Vi from the equation is 2.8 m/s.

The equation for determining the kinetic energy is:
KE = 0.5mv²

Substituting,
KE = 0.5(0.60 x 10^-3g)(1 kg/1000g)(2.8 m/s)²
KE = 2.352 x 10^-6 J

4 0

2 years ago

Read 2 more answers

In 1932 Albert Dremel of Racine, Wisconsin, created his rotary tool that has come to be known as a dremel.

kramer

Answer:

3054.32618 rad/s²

-431.1989 rad/s²

29080

Explanation:

Converting angular speed to rad/s

$\omega=35000\times \frac{2\pi}{60}=3665.19142\ rad/s$

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3665.19142-0}{1.2}\\\Rightarrow a=3054.32618\ rad/s^2$

The average acceleration while speeding up is 3054.32618 rad/s²

The number of turns in the 1.2 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 3054.32618\times 1.2^2\\\Rightarrow \theta=2199.11484\ rad=\frac{2199.11484}{2\pi}=349.99\ rotations$

The number of rotations in the 1.2 seconds is 349.99

Number of rotations in the 45 seconds

$\frac{35000}{60}\times 45=26250\ rotations$

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3665.19142}{8.5}\\\Rightarrow a=-431.1989\ rad/s^2$

Average angular acceleration while slowing down -431.1989 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3665.19142^2}{2\times -431.1989}\\\Rightarrow \theta=15577.0668\ rad=\frac{15577.0668}{2\pi}\\ =2479.16718\ rotations$

Number of rotations while slowing down is 2479.16718

Total number of rotations is 349.99+26250+2479.16718 = 29079.15718 = 29080

3 0

2 years ago

A 6 kg brick is pulled across the flat sidewalk by a horizontal force of 35 N. The brick is experiencing a frictional

SIZIF [17.4K]

<h3>Answer: Approximately 4.67 m/s^2</h3>

==============================================

Explanation:

Let's say you want to push the brick to the right. The free body diagram will have an arrow pointing right on the rectangle (the brick) and the arrow is labeled with 35 N.

Friction always counteracts whatever force you apply. The friction force arrow will point left and be labeled with 7 N.

The net horizontal force is therefore 35-7 = 28 N and the direction is to the right. The positive net force means you've overcome the force of friction and the brick is moving.

F = 28 is the net force

m = 6 is the mass

a = unknown acceleration

F = m*a .... newton's second law

28 = 6a

6a = 28

a = 28/6

a = 4.67

The acceleration of the brick is approximately 4.67 m/s^2

This means that for every second, the brick's velocity is increasing by about 4.67 m/s.

6 0

2 years ago