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vladimir1956 [14]
3 years ago
6

Charge q is 1 unit of distance away from the source charge S. Charge p is four times further away. The force exerted between S a

nd q is _____ the force exerted between S and p.
one0fourth
four times
one0sixteenth
sixteen times
Physics
2 answers:
velikii [3]3 years ago
7 0
The correct answer for the question that is being presented above is this one: "one0fourth." Charge q is 1 unit of distance away from the source charge S. Charge p is four times further away. The force exerted between S and q is one0fourth the force
zhenek [66]3 years ago
4 0

The correct answer is:

sixteen times


In fact, the distance between charge q and the source S is 1 unit. Instead, the distance between charge p and the source S is 4 units. The magnitude of the electrostatic force is inversely proportional to the square of the distance between the charge and the source:

E \sim \frac{1}{r^2}

where r is the distance. If we take the force between q and S as reference, we have r=1, so that

E_{qS}=\frac{1}{1^2}=1

while the force between p and S is

E_{pS}=\frac{1}{4^2}=\frac{1}{16}

Therefore, we see that the force exerted between q and S is 16 times the force exerted between p and S.

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Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L
snow_lady [41]

Answer:

Q=81.56\ W/m^2

Explanation:

Given that

T_1= 210 K

T_2= 150 K

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

Q=\sigma (T_1^4-T_2^4)\ W/m^2

So now by putting the values

Q=\sigma (T_1^4-T_2^4)\ W/m^2

Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2

Q=81.56\ W/m^2

So rate of heat transfer per unit area

Q=81.56\ W/m^2

8 0
3 years ago
A spherical drop of water carrying a charge of 43 pC has a potential of 540 V at its surface (with V = 0 at infinity). (a) What
almond37 [142]

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V  

(a) The electric potential on the surface is given by :

V=\dfrac{kq}{r}

r is the radius of the drop

r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

\dfrac{4}{3}\pi r^3+\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi R^3\\\\2r^3=R^3\\\\R=2^{1/3} r

Now the charge on the new drop is 2q. New potential is given by :

V=\dfrac{9\times 10^9\times 43\times 10^{-12}\times 2}{2^{1/3}\times 7.17\times 10^{-4}}\\\\V=856.79\ V

Hence, the radius of the drop is 7.17\times 10^{-4}\ m and the potential at the surface of the new drop is 856.79 V.

5 0
3 years ago
A wooden block floats a liquid with 1/4 of it's volume above the liquid.find the density of the liquid given that the density of
exis [7]

675g/cm³ is the density of the liquid

Here we have given that

density of wood be ρ.

the volume of the block be V          

The volume of the block =1/4 v

The volume of the immersed block (v) =V-1/4V=3/4V

We know the weight of the block = the weight of the water displaced by the submerged part of the block.

i.e. V x ρ x g    

= 3/4 x900g/cm³                            

= 675g/cm³

Density is a measure of mass divided by the volume of an object. Defined as mass per unit volume. here from the above equation we have substituted the given value and  we got the answer as 675g/cm³

       

Learn more about Density here brainly.com/question/1354972

 

#SPJ9

7 0
1 year ago
A +5.00 pC charge is located on a sheet of paper.
emmainna [20.7K]

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = q_{int} /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

8 0
3 years ago
9. From the densities below, decide whether each substance will sink or float in sea water. Sea water
tekilochka [14]

Answer:

Gasoline will float

Asphalt will sink

Cork will float

Explanation:

Simply compare the value of each object's density to that of the sea water (1.025 g/ml). If the density of the object is less than that of the water, the object will float due to the buoyance force.

Contrarily, if the density of the object is larger than that of sea water, the object will sink.

Gasoline, with density 0.66 g/ml which is less than that of sea water, will float.

Gasoline, with density 1.2 g/ml which is more than that of sea water, will sink.

Cork, with density 0.26 g/ml which is less than that of sea water, will float.

8 0
3 years ago
Read 2 more answers
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