Ask question

Ask question

All categories

3 years ago

8

The rectangular region of the xy plane shown has nonuniform surface charge density σ = σ0 (1+ y b), where σ0 is a constant. Find

the net charge.

1 answer:

sergiy2304 [10]3 years ago

8 0

Answer:

This is net charge on the surface is Q = σ₀ x (y + 2by²)

Explanation:

The surface charge density is defined as the amount of charge Q per unit area A

σ = dq / dA

dq = σ dA

Since the surface is a rectangular region we use an xy coordinate system so the area difference

dA = dxdy

dq = σ dx dy

We replace, evaluate the integral

∫ dq = ∫ σ₀ (1 + yb) dxdy

realizamos laintegral de dx

Q -0 =σ₀ ∫ (1 + yb) (x-0) dy

Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0

We perform the other integral (dy)

Q = σ₀ x (y + 2y² b)

Evaluated between Y = 0 and Y = y

Q = σ₀ x (y + 2by²)

This is net charge on the surface

You might be interested in

When do sunspots disappear?

dedylja [7]

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

once the magnetic field weakens the area and cold plasma enters the area of the sunspot

8 0

3 years ago

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

3 years ago

What is the potential energy of a 30 Newton ball that is on the ground

Juli2301 [7.4K]

The potential energy of a 30N ball on the ground will be zero. With respect to height, h. Potential energy will be calculated like this. P=mgh. So if its on the ground relatively speaking the h=0. Thus inputting into the above formula. P=0.

3 0

3 years ago

a 45 kg ice skater initially skating at a velocity of 3 m/s speeds up to a velocity of 5 m/s. calculate the difference in the ma

ad-work [718]

Answer: 90 kgm/s

Explanation:

The momentum (linear momentum) $p$ is given by the following equation:

$p=m.V$

Where:

$m=45 kg$ is the mass of the skater

$V$ is the velocity

In this situation the skater has two values of momentum:

Initial momentum: $p_{1}=m.V_{1}$

Final momentum: $p_{2}=m.V_{2}$

Where:

$V_{1}=3 m/s$

$V_{1}=5 m/s$

So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):

$p=p_{2}-p_{1}=m.V_{2}-m.V_{1}$

$p=m(V_{2}-V_{1})$

$p=45 kg(5 m/s - 3 m/s)$

Finally:

$p=90 kgm/s$

4 0

3 years ago

Read 2 more answers

We can determine the velocity of a wave when given the _______ and the wavelength.

ehidna [41]

Answer:

velocity = frequency * wavelength

Given the wavelength one must also have the frequency.

7 0

3 years ago