Which characteristics describe a point charge

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

Lesechka [4]

Answer:

$\Delta V = 0.053 A$

Explanation:

Electric field in a given region is given by equation

$E = \frac{A}{r^4}$

as we know the relation between electric field and potential difference is given as

$\Delta V = -\int E. dr$

so here we have

$\Delta V = - \int (\frac{A}{r^4}) .dr$

$\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}$

here we know that

$r_1 = 1.71 m$ and $r_2 = 2.89 m$

so we will have

$\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})$

so we will have

$\Delta V = 0.053 A$

8 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago

What is the name of the device that regulates the amount of fuel going into the engine of a personal watercraft (pwc)?

defon

The throttle is the name of the device that regulates the amount of fuel going into the engine of a personal watercraft (PWC). The throttle make the PWCs maneuverable. TO MAINTAIN STEERAGE, YOU MUST APPLY THROTTLE!<span> They are built for quick, sharp turns, low-radius circling, and rapid acceleration</span>

4 0

3 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m

Anna35 [415]

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

$I = \frac{P}{4\pi r^2},$

Where

P = Power

r = Radius

Replacing the values that we have,

$I = \frac{60}{(4*\pi (0.7)^2)}$

$I = 9.75 Watt/m^2$

The definition of intensity tells us that,

$I = \frac{1}{2}\frac{B_o^2 c}{\mu}$

Where,

$B_0 =$ Magnetic field

$\mu =$ Permeability constant

c = Speed velocity

Then replacing with our values we have,

$9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}$

Re-arrange to find the magnetic Field B_0

$B_o = 2.86*10^{-7} T$

Therefore the amplitude of the magnetic field of this light is $B_o = 2.86*10^{-7} T$

6 0

3 years ago

When do vernal equinoxes usually occur? autumnal?

shepuryov [24]

Spring tides occur twice each lunar month all year long without regard to the season. Neap tides, which also occur twice a month, happen when the sun and moon are at right angles to each other. ... The moon appears full when the Earth is between the moon and the sun

7 0

3 years ago