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crimeas [40]
2 years ago
12

A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C

/m^3. The outer layer has a uniform charge density of +8.0C/m^3 and extends from an inner radius of 6.0 cm to an outer radius of 12.0 cm. Find the total charge Q on the sphere.
Physics
1 answer:
rosijanka [135]2 years ago
3 0

Answer:

The total charge on the sphere is 46.11 x 10⁻³ C

Explanation:

Given:

charge density of innermost section, σ = −5.0C/m³

radius of the innermost section, r = 6.0cm

charge density of the outer layer ,σ = +8.0 C/m³

radius of the outer layer, r =12.0cm

volume of sphere is given as = \frac{4}{3}.\pi  r^3

Charge enclosed by the innermost section = volume x charge density

volume = \frac{4}{3}\pi  r^3 = \frac{4}{3}\pi  (0.06^3) = 9.05  X 10^{-4} m^3

Enclosed charge, q₁ = −5.0C/m³ X 9.05 x 10⁻⁴ m³

                             = - 4.53 x 10⁻³ C

Charge at the outer surface

Volume =  \frac{4}{3}\pi  [r_2{^3} - r_1{^3}] = \frac{4}{3}\pi  [0.12{^3} - 0.06{^3}] = 0.00633 {m^3}

Enclosed charge, q₂ = +8.0 C/m³ X 0.00633 m³

                            = 50.64 x 10⁻³ C

Total charge on the sphere; Q = q₁ + q₂

                                                   = - 4.53 x 10⁻³ C +  50.64 x 10⁻³ C

                                                   =  46.11 x 10⁻³ C

Therefore, the total charge on the sphere is 46.11 x 10⁻³ C

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Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

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6 0
3 years ago
A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070
likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

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Reil [10]

\LARGE{ \underline{\underline{ \purple{ \bf{Required \: answer:}}}}}

GiveN:

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Using the first Equation of motion,

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<h3>Note:</h3>

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