A car travelling at 10 m/s accelerates at 3 m/s^2 for 6 seconds. What is the final velocity of the car?

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

2 years ago

Yeast breaking down starch into usable energy ____________ law of thermodynamics.

Anon25 [30]

<span>First law of thermodynamics. This conservation law states that energy cannot be created or destroyed but can be changed from one form to another. In essence, energy is always conserved but can be converted from one form into another. Like when an engine burns fuel, it converts the energy stored in the fuel's chemical bonds into useful mechanical energy and then into heat, or more specifically, the melting ice cubes. Yeast breaks down maltose into glucose to produce alcohol and Co2 in the fermentation process. This is a prime example of the 1st law of thermodynamics. No form of usable energy is really lost; it only changes from one form to another</span>

4 0

3 years ago

Read 2 more answers

Write down an equation describing a sinusoidal traveling wave (in 1-D). Tell us (words and/or equations) what in your equation t

Natalija [7]

Answer:

Explanation:

The standard equation of the sinusoidal wave in one dimension is given by

$y = A Sin\left ( \frac{2\pi }{\lambda }\left ( vt-x \right )+\phi \right )$

Here, A be the amplitude of the wave

λ be the wavelength of the wave

v be the velocity of the wave

Φ be the phase angle

x be the position of the wave

t be the time

this wave is travelling along positive direction of X axis

The frequency of wave is f which relates with velocity and wavelength as given below

v = f x λ

The relation between the time period and the frequency is

f = 1 / T.

6 0

3 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$

$\frac{1}{f_2}$ = 0.06

f₂ = 16.67 cm

the expression for the power of the lenses is

P = $\frac{1}{f}$

where the focal length is in meters

1) P₁ = 1/0.50

P₁ = -2 D

2) P₂ = 1 /0.16667

P₂ = 6 D

4 0

2 years ago

What is the mass of an object that is hanging 12.6 m above the surface of the earth and has a

satela [25.4K]

Answer:

22.05 Kg

Explanation:

Apply the formula:

GPE = Gravity . Mass . ΔHigh

2778.3 = 10 . Mass . 12.6

2778.3 = 126 . Mass

Mass = 2778.3/126

Mass = 22.05

7 0

3 years ago