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3 years ago

Which of the following are examples of etiquette? Check all of the boxes that apply.

Physics

1 answer:

Irina18 [472]3 years ago

6 0

Answer:

C& D

Explanation:

ON EDG2020

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Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the tot

guapka [62]

Answer:

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

$E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}$

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

$R = \frac{d}{\sqrt2}$

now we have

$E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}$

so the force on the charge is given as

$F = QE$

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

3 0

3 years ago

What are two ways electromagnetic waves are used in a home computer scanner?

miss Akunina [59]

Answer:

A and B

Explanation:

8 0

2 years ago

Read 2 more answers

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

3 years ago

A man jogs at a speed of 1.6 m/s. His dog

FromTheMoon [43]

I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m

5 0

3 years ago

What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?

sveta [45]

Momentum = (mv).
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>

8 0

3 years ago