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3 years ago

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The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of Q is located at one of the corn

ers of the square. What is the potential (relative to infinity) at the center when each of the other corners is also contains a point charge of Q

1 answer:

tester [92]3 years ago

5 0

Answer:

12.0 V

Explanation:

Data :

Potential difference due to a single charge (+Q), E = 3.0 V

The Electric potential for the system of charges is given as:

$E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}]$

for single charge, E = 3.0 V = $\frac{1}{4\pi \epsilon_o}[\frac{Q}{r}]$ ->eq(1)

And for 4 charges:

$E=\frac{1}{4\pi \epsilon_o}[4\frac{Q}{r}]$ -eq(2)

from eq(1) and (2) we have

E = 4 × 3.0 V = 12 V

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A. Obtain the following: microwave, ruler, something meltable (e.g. candy bar, marshmallows) B. A microwave works by setting up

Makovka662 [10]

Answer:

E = 124.7 N / C

Explanation:

Let's analyze the exercise: the microwave creates an electromagnetic wave of frequency F = 2.45 GHz, this wave is introduced into the microwave cavity and is reflected on the metal walls, which is why one or more standing waves are formed.

The electric field of the standing wave is

I = E²

E =√I

where I is the intensity of the radiation.

What is it

I = P / A

where P is the effective emission power, almost all the power of the microwave and A is the area of the cavity, in the most used microwaves

P = 700 W and the area is A = 25 x 18 cm² = 0.045 m²

I = 700 / 0.045

I = 15555.56 W/m²

let's calculate the electric field

E = √15555.56

E = 124.7 N / C

7 0

3 years ago

A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten

Lyrx [107]

Answer:

a) the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ $_{max}$ = 16T $_{max}$ /πd³

Therefore, the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta ( $\theta$ ) be the angle of twist

polar moment of inertia $I_p}$ = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d $\theta$ = T(x)dx / G $I_{p}$

so

d $\theta$ = tx.dx / G(πd⁴/32)

d $\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ = $\int\limits^L_0 \, d\theta$

$\theta$ = $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴ $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴

Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴

7 0

3 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

2 years ago

A plastic box has an initial volume of 2.00 m 3 . It is then submerged below the surface of a liquid and its volume decreases to

nikitadnepr [17]

Answer:

Volume strain is 0.02

Explanation:

Volume strain is defined as the change in volume to the original volume.

It is given that,

Initial volume of the plastic box is 2 m³

It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³

We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,

$\delta V=\dfrac{V_f-V_i}{V_i}\\\\\delta V=\dfrac{1.96-2}{2}\\\\\delta V=0.02$

So, the volume strain on the box is 0.02.

6 0

3 years ago