Ask question

Ask question

All categories

3 years ago

12

A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field = (100 N/C) î. Calculat

e the flux of the net electric field through a Gaussian sphere of radius R = 10.0 cm centered at the origin. (ε0 = 8.85 × 10-12 C2/N ∙ m2)

1 answer:

SVETLANKA909090 [29]3 years ago

3 0

Answer:

$\phi = 2.26 \times 10^5 Nm^2/C$

Explanation:

Due to uniform electric field net flux through a closed surface is always zero

so here we have to find the net flux through a sphere due to a point charge placed at origin and due to a uniform electric field

so here we will have

$\phi = \frac{q}{\epsilon_o}$

now we have

$\phi = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}}$

$\phi = 2.26 \times 10^5 Nm^2/C$

Due to external uniform field the electric flux is zero

so the total flux is only due to charge and it is equal to

$\phi = 2.26 \times 10^5 Nm^2/C$

You might be interested in

Your friend comes back to school after a week of being sick.

Serhud [2]

That Organ is be percussion.

6 0

3 years ago

A 0.20 kg mass (m1) hangs vertically from a spring and an elongation of the spring of 9.50 cm (r1) is recorded. With a mass (m2)

kupik [55]

Answer:

k=320N/m

Explanation:

Step one:

given data

Let the initial/equilibrum position be x

mass m1= 0.2kg

F1= 0.2*10= 2N

elongation e= 9.5cm= 0.095m

mass m2=1kg

F2=1*10= 10N

elongation e= 12cm= 0.12m

Step two:

From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F

F=ke

2=k(0.095-a)

2=0.095k-ka----------1

10=k(0.12-a)

10=0.12k-ka----------2

solving equation 1 and 2 simultaneously

10=0.12k-ka----------2

- 2=0.095k-ka----------1

8=0.025k-0

divide both side by 0.025

k=8/0.025

k=320N/m

5 0

3 years ago

Use the drop-down menus to complete each sentence about the layers of the atmosphere. If the did not exist, Earth might be destr

ozzi

Answer:

If the mesosphere did not exist, Earth might be destroyed by chunks of rock from space.

The stratosphere is located 12 to 50 kilometers from Earth's surface.

Both the trophosphere and mesosphere get colder as altitude increases.

The ozone in the stratosphere protects people from ultraviolet (UV) radiation.

The thermosphere has the highest temperature of any layer in Earth's atmosphere.

Explanation:

Different layers of the atmosphere are;

Troposphere

It is the lowest layer of atmosphere that begins at ground level and extends upward to about 10 km. In this layer, all organisms live, weather occurs and most clouds appear. As altitude increases, the temperature decreases i.e., the layer gets colder.

Stratosphere

This layer is located at the top of the troposphere and extends up to about 50 km from Earth's surface. The ozone layer present in this layer absorbs high-energy ultraviolet (UV) light from the Sun. Here, the temperature rises with altitude. The passenger jets are usually fly in this layer.

Mesosphere

Above the stratosphere, the mesosphere extends upward to about 85 km from Earth's surface. The chunks of rock from space such as meteors burn upon entering the mesosphere and protect the Earth. In this layer, the temperature gets colder as altitude increases and so the coldest temperatures (about -90° C) in Earth's atmosphere are found at the top of this layer.

Thermosphere

This layer is located above the mesosphere and the air here is really very thin. The thermosphere absorbs high-energy X-rays and UV radiation from the Sun and as a result, its temperature increases to hundreds or thousands of degrees i.e., 1,500° C or higher. This layer extends up to 500 to 1,000 km from the ground. Many satellites orbit Earth within this layer.

Ionosphere

The ionosphere is actually an extension of the thermosphere and so it is not a distinct layer like the others. It is filled with charged particles produced from the ionization of molecules by high temperatures in the thermosphere caused by the high-energy radiation from the Sun. The phenomenon called auroras (Northern Lights and Southern Lights) occurs in this layer. The different regions of the ionosphere reflecting the radio waves back to Earth and make the long-distance radio communication possible.

4 0

3 years ago

Read 2 more answers

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

1:What kind of energy does your body use to keep you warm.

Zigmanuir [339]

Answer:

1. thermal energy

2. Batteries and coal

4. energy conversion

3 0

3 years ago