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natali 33 [55]
3 years ago
12

A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ositive test charge (q0 = +4.56 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -6.95 x 10-9 J. Find rB.
Physics
1 answer:
ella [17]3 years ago
7 0

Answer:

r_B = 1.88 m

Explanation:

As we know that work done by electric force is given as

W_e = -q\Delta V

so here we know that charge is moving from

r_A = 2.97 m

to another position

so we will have

W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}

-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})

-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})

r_B = 1.88 m

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How many total oxygen atoms are in the compound Molybdenum (V) Dichromate?
alekssr [168]

Answer:

1.2646\times10^{25}\ atoms

Explanation:

-The chemical formula for Molybdenum (V) Dichromate is Mo(Cr_2O_7)_3

-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate

-We apply Avogadro's constant to find the number of atoms of oxygen:

Avogadro's \ Constant=6.022\times 10^{23} \ mol_1\\\\No\ of \ Atoms=Moles\times Avogadro's \ Constant\\\\=21\times 6.022\times 10^{23} \\\\=1.2646\times10^{25}\ atoms

Hence, there are 1.2646\times10^{25} \ atoms

3 0
3 years ago
Use the equation for magnetic force on a moving charge to derive the equation for magnetic force on a current carrying wire. Sho
max2010maxim [7]

Answer:

The formula comes from Lorentz force law which includes both the electric and magnetic field. If the electric field is zero, the force law for just the magnetic field is <u>F=q(ν×B</u>) . Here, F  is force and is a vector because the force acts in a direction.  q  is the charge of the particle.  v  is velocity and is a vector because the particle is moving in some direction.  B is the magnetic flux density.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd.  Since the magnitude of B is constant at every line element of the loop (circle) and it dot product with the line element is B dl everywhere, therefore

                                                  ∮B dl=μ0 I

                                                  B ∮dl=μ0 I

                                                  B 2πr=μ0 I

                                                   B=μ02πr Id=μ0/4π I dl×rr3

Since, r can be written as r=(rcosθ,rsinθ,z) and dl as dl=(dl,0,0) And now, if we take the cross product we would get

                                               dl×r=−z dlj^+rsinθk^

and therefore the magnitude of dB is equal to

dB=μ0/4π I |dl×r|/r3=μ0/4π I z2+r2sin2θ−−−−−−−−−−√dl/r3

Thus, magnetic field is depending on r,θ,z.

Learn more about Force here-

brainly.com/question/2855467

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7 0
1 year ago
If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the
kodGreya [7K]

Answer:

a)

v = 4.048 *10^6 m/s

b)  

Angular frequency =  1.92 * 10^7

Explanation:

As we know

v =  \frac{qBr}{m}

q is the charge on the electron = 3.2 * 10^{-19} C

B is the magnetic field in Tesla = 0.4 T

r is the radius of the circle = 0.21 m

mass of the electrons = 6.64 * 10^{-27} Kg

a)

Substituting the given values in above equation, we get -

v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6m/s

b)  

Angular frequency =

\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7

8 0
2 years ago
A force of 20 N acts upon a 5 kg block. What is the acceleration of the object.
horsena [70]

Answer:

4m/s/s

Explanation:

a=f/m

m=5kg

f=20N

20/5=4

(N=kg-m/s/s)

7 0
2 years ago
Read 2 more answers
In the diagram, R1 = 40.0 ,
Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0
2 years ago
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