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Sergio039 [100]
3 years ago
5

A solid conducting sphere of radius 2.00 cm has a charge of 6.77 μC. A conducting spherical shell of inner radius 4.00 cm and ou

ter radius 5.00 cm is concentric with the solid sphere and has a charge of −2.13 μC. Find the electric field at the following radii from the center of this charge configuration. (a) r = 1.00 cm (b) r = 3.00 cm (c) r = 4.50 cm (d) r = 7.00 cm
Physics
1 answer:
madreJ [45]3 years ago
5 0

Answer:

Part a)

E = 0

Part b)

E = 6.77 \times 10^7 N/C

Part c)

Electric field inside the conductor is again zero

E = 0

Part d)

E = 8.52 \times 10^6 N/C

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

E = \frac{kq}{r^2}

E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}

E = 6.77 \times 10^7 N/C

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

E = 0

Part d)

Now at r = 7 cm

again by Gauss law

E = \frac{kQ}{r^2}

E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}

E = 8.52 \times 10^6 N/C

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blondinia [14]

Answer:

76969.29 W

Explanation:

Applying,

P = F×v............. Equation 1

Where P = Power, F = force, v = velocity

But,

F = ma.......... Equation 2

Where m = mass, a = acceleration

Also,

a = (v-u)/t......... Equation 3

Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s

Substitute these values into equation 3

a = (12.87-0)/3.47

a = 3.71 m/s²

Also Given: m = 1612 kg

Substitute into equation 2

F = 1612(3.71)

F = 5980.52 N.

Finally,

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3 0
2 years ago
What causes atoms to bond together<br> ?!?
lisov135 [29]

Answer:

electrostatic attraction

Explanation:

Atoms form chemical bonds with other atoms when there's an electrostatic attraction between them. This attraction results from the properties and characteristics of the atoms' outermost electrons, which are known as valence electrons.

7 0
3 years ago
Read 2 more answers
A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.
UkoKoshka [18]

Answer:

a) 1.6 mN  b) -1.6 mN  c) -1.6 mN  d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the  line that joins the  charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ =  1.6 mN (going in the positive x direction, towards q₂)

6 0
3 years ago
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1. A baseball is thrown vertically at 16.7 m/s. What is the maimum height of the baseball?
Anvisha [2.4K]

Answer:

14.2 m

Explanation:

Using conservation of energy:

PE at top = KE at bottom

mgh = ½ mv²

h = v² / (2g)

h = (16.7 m/s)² / (2 × 9.8 m/s²)

h = 14.2 m

Using kinematics:

Given:

v₀ = 16.7 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 14.2 m

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3 years ago
An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of
slega [8]

Answer:

T_{1}=94.9^{o}C

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

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The temperature of heater surface follows from the rate equation written as:

T_{1}=T_{2}+\frac{q}{kS}

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\

Substitute the given values

S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m

The temperature of heater is then:

T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

                           T_{1}=94.9^{o}C

5 0
2 years ago
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