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inessss [21]
3 years ago
11

A solid, cylindrical wire conductor has radius R = 30 cm. The wire carries a current of 2.0 A which is uniformly distributed ove

r the cross-section of the wire (current density is constant). What is the magnitude of the magnetic field due to the current in the wire at a radial distance of r = 200 cm from the center axis of the wire? HINT: Use Ampere’s law, noting that B is tangential.
Physics
1 answer:
blondinia [14]3 years ago
7 0

Answer:

Explanation:

The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.

Applying Ampere's law :-

∫ B dl = μ₀ I      ; I is current passing through ampere's loop

B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ T.

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A simple pendulum oscillates back and forth with a maximum angular displacement of pi/12 radians. At t=1.2s, the pendulum is at
Zielflug [23.3K]

Answer:

a)  w = 1.08 rad / s , b) θ = π / 12 cos (1.08 t - 1,296) , c)  w = - 0.2827 sin (1.08t - 1,296) , d) energy decrease

Explanation:

a) The oscillatory movement of a simple pendulum has the angular velocity

      w = √(L / g)

      w = √(0.35 / 0.300)

      w = 1.08 rad / s

b) it is requested to find the equation of angular displacement

       θ = θ₀ cos (wt + φ)

We replace

       θ = π / 12 cos (1.08 t + φ)

To find the constant let's use the value they give

       θ = θ₀     for   t = 1.2 s

       θ = θ₀ cos 1.08 1.2 + φ)

       Cos (1,296 + φ) = 1

        1,296 + φ = cos⁻¹ 1  

Remember that the angles must be in radians

         φ = 0 - 1,296

The final equation is

        θ = π / 12 cos (1.08 t - 1,296)

c) the angular velocity is

         w = dθ / dt

         w = π / 12 (- 1.08 sin (1.08t - 1,296))

         w = - 0.2827 sin (1.08t - 1,296)

d) if addamping force is included, part of the energy dissipates, therefore the total energy must decrease

e) The period of a physical pendulum is

      w = √ (mg d / I)

Where I is the moment of inertia that is of the form

      I = cte m R²

     w = √ (m g d / cte md2) = √ (g / d)  √(1/cte)

Simple pendulum wo = √ (g / d)

       w = w₀  1/√cont

As we see the angular velocity of this pendulum change due to the constant that accompanies the moment of inertia. In general this constant is less than by which the angular velocity the angular velocity increases

6 0
3 years ago
A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
irina [24]

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
  • For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
  • An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

4 0
2 years ago
Several objects were dropped from a balcony. Using the data below, determine which statements must be true regarding the rate of
NeX [460]

Answer:

statement three and statement 2

Explanation

since they are weighted different, they will have more or less force (more mass=more force), also the heavier an object, the greater its acceleration while falling.

6 0
3 years ago
An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a con
antoniya [11.8K]

(a) 4.03 s

The initial angular velocity of the wheel is

\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s

The angular acceleration of the wheel is

\alpha = -3.40 rad/s^2

negative since it is a deceleration.

The angular acceleration can be also written as

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 0 is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

\theta= \omega_i t + \frac{1}{2}\alpha t^2

where we have

\omega_i=13.7 rad/s is the initial angular velocity

\alpha = -3.40 rad/s^2 is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad

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3 years ago
Which item has the most potential energy
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