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3 years ago

14

A charge of 4 nc is placed uniformly on a square sheet of nonconducting material of side 17 cm in the yz plane. (a) what is the

surface charge density sigma ?

1 answer:

Ratling [72]3 years ago

8 0

The charge density of the sheet is 1.384×10⁻⁷C/m².

Charge density is defined as the charge per unit area.

The sheet is a square of length l=17 cm.

Calculate the area A of the sheet .

$A=l^2=(17 cm)^2= (17*10^-^2m)^2=0.0289 m^2$

The charge Q on the sheet is

$Q=4nC=4*10^-^9C$

The charge density σ is given by,

$\sigma=\frac{Q}{A}$

Substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

$\sigma=\frac{Q}{A}\\ =\frac{4*10^-^9C}{0.0289 m^2} \\ =1.389*10^-^7C/m^2$

Thus, the charge density of the sheet is <u>1.384×10⁻⁷C/m².</u>

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Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

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

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

C)

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

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2 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

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

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

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3 years ago

Is it possible that a speed of 254 and a speed of 100 could be the same speed?

EastWind [94]

Not if both speeds are in the same units.

However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
then the speeds are equal.

4 0

3 years ago

Will give Brainliest!! Question attached.

ss7ja [257]

Answer:

im pretty sure it is 3.0 K

Explanation:

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2 years ago

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A body of volume 2000 cm3 has a mass 4 kg. Find the density of the body. Will the body sink or float in water, if the density of

Brums [2.3K]

Answer:

$2 g/cm^3$ , it will sink

Explanation:

The density of an object is given by

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

For the body in the problem, we have

m = 4 kg = 4000 g

$V=2000 cm^3$

Therefore, its density is

$d=\frac{4000}{2000}=2 g/cm^3$

And the object will sink in water, because its density is larger than that of water, which is $1 g/cm^3$ . (an object sinks when its density is larger than that of water, otherwise it floats).

6 0

3 years ago