Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of the square is 2a. What is the electric potential at P, the center of the square?a. kQ/4a
b. kQ/a
c. zero volts
d. 2kQ/a
e. 4kQ /a

Answers

Answer 1

Answer:

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

$l = 2a$

The distance of the center of the square from each of the corners is

$r = (2a)/(2)= a$

The potential electric at the center due to each cornet charge is

$V_1 = (kQ_1)/(r_1)$

$V_2 = (kQ_2)/(r_2)$

$V_3 = (kQ_3)/(r_3)$

$V_4 = (kQ_4)/(r_4)$

The total electric potential at the center of the given square is

$V = V_1+V_2+V_3+V_4$

$V = (kQ_1)/(r_1)+ (kQ_2)/(r_2)+(kQ_3)/(r_3)+(kQ_4)/(r_4)$

Al the charges are equal, and the distance are equal to a, then

$V = (kQ)/(a)+ (kQ)/(a)+(kQ)/(a)+(kQ)/(a)$

$V = (4kQ)/(a)$

Therefore the correct option is E.

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What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Answers

Answer:

The velocity at discharge is 100.46 ft/s

Explanation:

Given that,

Pressure = 68 psi

We need to calculate the pressure in pascal

$P=68*6894.74\ Pa$

$P=468842.32\ Pa$

We need to calculate the velocity

Let the velocity is v.

Using Bernoulli equation

$P=(1)/(2)\rho v^2$

$468842.32=0.5*1000* v^2$

$v=\sqrt{(468842.32)/(0.5*1000)}$

$v=30.62\ m/s$

Now, We will convert m/s to ft/s

$v =30.62*3.281$

$v=100.46\ ft/s$

Hence, The velocity at discharge is 100.46 ft/s

Final answer:

The speed of water discharged from a hose depends on the nozzle pressure and the constriction of the flow, but the specific speed cannot be determined from pressure alone without additional parameters.

Explanation:

The question is asking about the velocity or speed achieved by water when it is forced out of a hose with a nozzle pressure of 68 psi. To understand this, we need to know that the pressure within the hose is directly correlated with the speed of the water's exit. This is due to the constriction of the water flow by the nozzle, causing speed to increase.

However, the specific velocity at discharge can't be straightforwardly calculated from pressure alone without knowing more details, such as the dimensions of the hose and nozzle, and the properties of the fluid. Therefore, based on the provided information, a specific answer in ft/sec can't be given.

Learn more about Pressure and Velocity here:

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