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JulijaS [17]
3 years ago
7

) Explain why the foil is attracted at first by the charged rod. Consider any charge that exists in the uncharged foil. (You wil

l consider later interaction between the rod and foil in the last question. Just think about the first interaction when answering this question.) [
Physics
1 answer:
Arisa [49]3 years ago
5 0

Answer:

The attraction is due to the induced charge.

Explanation:

When we approach a charged rod to a sheet, an induced load is produced in the sheet that is of the same magnitude as the rod of opposite sign, this is because the charges of different sign attract each other, this explains the initial attraction.

This induced load occurs if importing the plate load

The attraction is due to the induced charge.

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A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripet
arlik [135]
The correct answer is 53 meters per second squared 
5 0
3 years ago
Read 2 more answers
A non uniform rod has mass
Doss [256]

Answer:

r_{cm} = L/3

Explanation:

Mass: M, Length: L.

\sigma (x) = b(L-x)

The formula that gives center of mass is

\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ...}{m_1 + m_2 + ...} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}

In the case of a non-uniform mass density, this formula converts to

\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx }

where the denominator is the total mass and the nominator is the mass times position of each point on the rod.

We have to integrate the mass density over the total rod in order to find the total mass. Likewise, we have to integrate the center of mass of each point (xσ(x)) over the total rod. And if we divide the integrated center of mass to the total mass, we find the center of mass of the rod:

\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx } = \frac{\int\limits^L_0 {xb(L-x)} \, dx }{\int\limits^L_0 {b(L-x)} \, dx } = \frac{b\int\limits^L_0{(xL - x^2)} \, dx }{b\int\limits^L_0 {(L-x)} \, dx } = \frac{\frac{x^2L}{2} - \frac{x^3}{3}}{Lx - \frac{x^2}{2}}\left \{ {{x=L} \atop {x=0}} \right.

Here x's are cancelled. Otherwise, the denominator would be zero.

r_{cm} = \frac{\frac{xL}{2}-\frac{x^2}{3}}{L-\frac{x}{2}}\left \{ {{x=L} \atop {x=0}} \right. = \frac{\frac{L^2}{2}-\frac{L^2}{3}}{L-\frac{L}{2}} = \frac{\frac{L^2}{6}}{\frac{L}{2}} = \frac{L}{3}

8 0
3 years ago
An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance
bagirrra123 [75]

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8 \frac{m}{sec}

Final velocity ( V ) = 0

Distance traveled before come to rest = 0.2667 m

Now use third law of motion V^{2} = u^{2} - 2 a s

Put all the values in above formula we get,

⇒ 0 = 0.8^{2} - 2 × a ×0.2667

⇒ a = 1.2 \frac{m}{sec^{2} }

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

5 0
4 years ago
If you have two linear polarizers whose extinction axis are 90 degrees relative to eachother, no light passes through the system
marin [14]

Answer:

The third polarizer can be placed midway between the first two polarizers with its extinction axis at 45° from either polarizer to maximize the amount of light that is transmitted (one-eight).

Explanation:

If light is incident on a polarizer, it allows only light that is parallel to its 'pass-through' axis to pass through untouched.

Light whose electric direction/vector is perpendicular to the 'pass through' axis will not pass through at all. Light whose electric direction/vector points in other directions (apart from those whose direction is parallel or perpendicular) passes through according to the magnitude of the component that is parallel to the 'pass-through' axis.

The polarizer blocks half of the incident light rays and the transmitted light is polarized in the direction of the 'pass-through' axis.

A new polarizer now place at a distance from the first polarizer with its 'pass-through' axis perpendicular to the first polarizer cancel out all the light that comes through from the first polarizer. Since the light electric vector needs to be parallel to the axis of the polarizer to pass through and all the parallelized light from the first polarizer are now incident perpendicularly to the axis of the second polarizer, no light rays pass through.

But, a third polarizer can be placed midway between the first two polarizers with its axis positioned at 45° from either polarizer. Thereby allowing exactly half of the light from the first polarizer to pass through. The explanation is just like that for the first one. (Light whose electric direction/vector points in other directions (apart from those whose direction is parallel or perpendicular) passes through according to the magnitude of the component that is parallel to the 'pass-through' axis).

Then the resultant from the middle polarizer reaches the initial second polarizer and half of the light is let through again. So that, at the end of the day, (1/2) × (1/2) × (1/2) of the initial incident ray is let through.

That is, to maximize the amount of light that is transmitted (one-eight of initial incident ray) the third Polaroid is place midway between the first two and at angle 45° to either one.

8 0
4 years ago
17. A volleyball weighs about 300 grams.
atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

  • 1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

  • \boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

  • m = 0.3 kg
  • g = 9.8 m/s²
  • h = 15 m
  • PE = ?

Use formula of potencial energy:

  • \boxed{\bold{PE=m*g*h}}

Replace and solve:

  • \boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}
  • \boxed{\boxed{\bold{PE=44.1\ J}}}

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0
4 years ago
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