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Nonamiya [84]
4 years ago
13

Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical v

alue +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
Physics
1 answer:
Brut [27]4 years ago
3 0

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 = \frac{2a}{2}= a

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

V_1 = \frac{kQ_1}{r_1}

V_2 = \frac{kQ_2}{r_2}

V_3 = \frac{kQ_3}{r_3}

V_4 = \frac{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 = \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}+\frac{kQ_3}{r_3}+\frac{kQ_4}{r_4}

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

V = \frac{kQ}{a}+ \frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a}

V = \frac{4kQ}{a}

Therefore the correct option is E.

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Density

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Two identical loudspeakers are driven in phase by a common oscillator at 750 Hz and face each other at a distance of 1.24 m. Loc
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Answer:

0.2286  m,  0.686 m and  1,143 m

therefore we see that there is respect even where the intensity is minimal

Explanation:

Destructive interference to the two speakers is described by the expression

       Δr = (2n +1) λ/2

where r is the distance,  λ the wavelength and n an integer indicating the order of the interference

let's locate the origin on the left speaker

let's find the wavelength with the equation

        v =  λ f

         λ = v / f

we substitute

      Δr = (2n + 1) v / 2f

let's calculate for difference values ​​of n

     Δr = (2n +1) 343/(2 750)

     Δr = (2n + 1) 0.2286

we locate the different values ​​for a minimum of interim

    n     Δr (m)

     0    0.2286

     1     0.686

     2    1,143

therefore we see that there is respect even where the intensity is minimal

6 0
3 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d
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Answer:

Option (D) is correct.

Explanation:

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

time taken= 30 s

The distance traveled is given by d= v t

d= 14 (30)

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Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

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Explanation:

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Transpiration: Moisture in the plants are absorbed by the roots and distributed in all the parts of plant. Moisture present in the underside of the leaves get converted into water vapors and releases in atmosphere

Evaporation : In this process , liquid state of water is converted into gaseous state of water that is water vapors.

Condensation: After evaporation , the water vapors rises up towards the sky. As they rise in the atmosphere the temperature decrease due which these water vapors stars getting converted into small water droplets. These small water drops combines with particles of dust , gas to form clouds

Precipitation: Water released from the clouds in the form of rain.

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