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nasty-shy [4]
3 years ago
9

A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 7 cm. The washer has a

uniform charge distribution and a total charge of 8 nC. What is the electric field on the axis of the washer at a distance of 33 cm from the center of the washer? (Enter the magnitude of the electric field.)
Physics
1 answer:
Kryger [21]3 years ago
6 0

Answer:

2583.9 N/C

Explanation:

Parameters given:

Outer diameter = 14 cm

Outer radius, R = 7cm = 0.07m

Inner diameter = 7 cm

Inner radius, r = 3.5 cm = 0.035m

Charge of washer = 8 nC = 8 * 10^(-9)C

Distance from washer, z = 33 cm = 0.33m

The electric field due to a washer (hollow disk) is given as:

E = k * σ * 2π [ 1 - z/(√(z² + R²)]

Where σ = charge per unit area

σ = q/π(R² - r²)

σ = 8 * 10^(-9) /(π*(0.07 - 0.035)²)

σ = 2.077 * 10^(-6) C/m²

=> E = 9 * 10^9 * 2.077 * 10^(-6) * 2π * [1 - 0.33/(√(0.33² + 0.07²)]

E = 117.467 * 10^3 * (1 - 0.978)

E = 117.467 * 10^3 * 0.022

E = 2583.9 N/C

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A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C
Rus_ich [418]

Answer:

1.50\ *10^{-6} }

Explanation:

Given

e=100 N/C

M=0.15 g

q=\ 22\  pC\\=\ 22\ *10^{-2}

The  ratio of the electric force on the bee to the bee's weight can be determined by the following formula

\frac{fe}{M*9.81}

\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}

=\ 1.50\ *10^{-6}

4 0
3 years ago
You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1
drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

  • <em>Mass of the pendulum, = M </em>
  • <em>Length of the pendulum, = L</em>
  • <em>Initial angular speed, </em>\omega _i<em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

I_i = \frac{1}{3} ML^2

The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0
2 years ago
An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

5 0
3 years ago
In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of
Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, a_{Net} = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, v_y = v × sin(θ)

∴ v_y = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - a_{Net} × t

∴ t = (v₁ - v₂)/a_{Net}  = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2·a_{Net}·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0
2 years ago
PLEASE HELP
Oksana_A [137]

Answer:

The answer is A

Explanation: hope this helps :)

5 0
2 years ago
Read 2 more answers
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