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STALIN [3.7K]
3 years ago
9

Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl

e of charge q as the particle travels from A to B is WAB and that done on the particle as it travels from A to C is WAC=−WAB/3.
a) How much electrostatic work is done on the particle as it travels from B to C?
Physics
1 answer:
Trava [24]3 years ago
4 0

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac =  q (  Va -  Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]  

= Wac - Wab

= Wac + 3Wac

= 4Wac

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An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat given to the system in thes
Mashcka [7]

Answer:

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6 0
2 years ago
A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a
Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0
2 years ago
Please answer this question
sergij07 [2.7K]

Explanation:

m = kg. v=m/s. g=m/s^2. h= m

>>1/2mv^2=mgh

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .

<u>>>kg m^2/s^2=kg m^2/s^2</u><u> </u>

<u>></u><u>></u><u>J</u>= J

3 0
3 years ago
INT Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of 12 pC; these are both very
horsena [70]

Answer:

W = 2.3 10² F_{e}

Explanation:

The force of the weight is

         W = m g

         

let's use the concept of density

         ρ= m / v

the volume of a sphere is

         V = \frac{4}{3} π r³

         V = \frac{4}{3} π (1.0 10⁻³)³

         V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

          m = ρ V

          m = 1000 4.1887 10⁻⁹

          m = 4.1887 10⁻⁶ kg

therefore the out of gravity is

          W = 4.1887 10⁻⁶ 9.8

          W = 41.05 10⁻⁶ N

now let's look for the electric force

           F_e = q E

           F_e = 12 10⁻¹² 15000

           F_e = 1.8 10⁻⁷ N

         

the relationship between these two quantities is

          \frac{W}{F_e} = 41.05 10⁻⁶ / 1.8 10⁻⁷

           \frac{W}{F_e} = 2,281 10²

             

             W = 2.3 10² F_{e}

therefore the weight of the drop is much greater than the electric force

3 0
3 years ago
What is the name of the covalent compound P4S2?
Roman55 [17]

Tetraphosphorus disulfide


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6 0
3 years ago
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