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3 years ago

Can someone help me on #18 - #26? At least one ☝️ I really need help!!!

Physics

1 answer:

Novosadov [1.4K]3 years ago

3 0

#18). (I think. It's the one that starts with "Compare..."
Gravity ALWAYS attracts.
The force between electric charges can attract or repel ...
it depends whether the charges are the same kind or opposite kinds.

#19).
With both gravity and electric charges, the force between them quickly
becomes weaker when the distance between them increases.

#20).
I don't think it changes.
If the doorknob gets charged by something that TOUCHES it, so that
charges can flow into it from the other object or out of it, then the total
amount of electric charge on it might change. But the question says
that the doorknob is charged by an "electric field", so nothing touched it,
and charges couldn't flow into it or out of it. The only way it got charged
was by the charges it already had in it getting moved around ... electrons
in one part of the knob moving over to the other side. Then it would act
as if it was charged ... if you touched it, you might get zapped.

#21)., #22)., #23).
You're supposed to draw a graph to answer these.
It's a very easy graph to draw, and you should do it.
Label the x-axis 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Label the y-axis 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2 .
Put the four points on the graph ... A, B, C, and D.
For each point, the 'battery voltage' is the number on the x-axis
and the 'Current' is the number on the y-axis.
Then draw a line through the points.
When you have the graph to look at, you can easily answer 21, 22, and 23.

24). I'm not sure, and I don't want to guess.

25). Did you ever move a coil of wire near a magnet in class ?
This is the same situation, only the magnet is moving and the wire is still.
The result will be the same.

26). Magnets have two poles that attract
the opposite kind and repel.
Now you copy the map and fill in the other side.

This is a lot of work for 5 points, so I left some parts for you to do.
Another reason I did that is: You'll learn a lot more that way.

You might be interested in

What is the magnitude and direction (right or left) of the

Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

In y-component:

$Fy_{net}=F_{n}+F_{g}$ (1)

Where $F_{n}=12 N$ is the Normal force, directed upwards and $F_{g}=-12 N$ is the weight of the box (gravity force), directed downwards.

$Fy_{net}=12 N-12 N$ (2)

$Fy_{net}=0 N$ (3) Hence the net force in the vertical component is zero

In x-component:

$Fx_{net}=F_{left}+F_{right}$ (4)

Where $F_{left}=-3 N$ and $F_{right}= 15 N$

$Fx_{net}=-3 N + 15 N$ (5)

$Fx_{net}=12 N$ (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0

3 years ago

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0

3 years ago

a vertical polarizing filter is used on the lens of a camera. Which of the following do not strike the lens?

ziro4ka [17]

A vertical polarizing filter is used on the lens of a camera, they block out the light that is horizontally polarized, so they allow all of the vertically polarized light to pass through.

8 0

3 years ago

After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.00with the original di

PSYCHO15rus [73]

Answer:

143 °

Explanation:

a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then

d sinθ = ( 2n+1) λ/2

for first dark fringe

d sinθ = λ/2

d /λ = 1/ 2 sinθ

1 / 2 sin15

= 1.93

b )

For intensity of fringe at angle θ, the relation is

I = I₀ cos²θ

I / I₀ = cos²θ/2

Given I / I₀ =0. 1

0.1 = cos²θ/2

θ/2 = 71.5

θ = 143 °

4 0

3 years ago

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$