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2 years ago

14

3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P

hoto 2.

1 answer:

KIM [24]2 years ago

6 0

Answer:

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

$E=\frac{F}{Q}$

But the force F

$F= \frac{kQ1Q2}{r^2}$

But the electric field intensity due to a point charge Q at a distance r meters away is given by

$E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }$

<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>

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A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4

HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$2.8 \text{ m} = 0+\dfrac{gt^2}{2}$

$t = \sqrt{\dfrac{5.6}{g}}$

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g </em> (It is going up)

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$1.2\text{ m} = ut-\frac{1}{2}gt^2$

Substituting the value of <em>t</em> from the previous equation,

$1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}$

$u\sqrt{\dfrac{5.6}{g}} = 4.0$

Taking <em>g</em> = 9.8 m/s²,

$u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}$

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

$u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}$

The initial velocity is

$v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}$

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2 years ago

Which of these discoveries contradict components of Dalton’s atomic theory? Check all of the boxes that apply. Atoms contain sma

EastWind [94]

Explanation:

According to Dalton's atomic theory, all the atoms are individual, all the atoms of the same element are identical in properties and mass, the compound is formed from two or more kinds of the atoms, all the matter is made up of small atoms and the chemical reaction is a rearrangement of the atoms.

The discoveries which contradicts the components of Dalton's atomic theory from the given discoveries are:

Nuclear reactions can change an atom of one element into an atom of another element.

Atoms of a given element can have different numbers of neutrons.

Atoms contain smaller particles: protons, neutrons, and electrons.

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2 years ago

Read 2 more answers

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

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2 years ago

As the drawing illustrates, a siren can be made by blowing a jet of air through 20 equally spaced holes in a rotating disk. The

Aneli [31]

Answer:

ω = 630.2663 = 630[rad/s]

Explanation:

Solution:

- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.

1rev(20 hole) -> 20(cycle)/rev

2006.2(cycle) -> f ?

f = 2006.2/20 = 100.31rev at second

- The relation between angular frequency and angular speed is given by:

ω = 2πf

ω = 2*3.14*100.31

ω = 630.2663 = 630[rad/s]

4 0

2 years ago

Батискаф витримує тиск 60 МПа. Чи можна провести дослідження

Elanso [62]

1) Yes

2) $6.34\cdot 10^9 N$

Explanation:

1)

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

$p=p_0+\rho g h$

where

$p_0 = 101,300 Pa$ is the atmospheric pressure

$\rho$ is the fluid density

$g=10 m/s^2$ is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

$\rho=1030 kg/m^3$ is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

$p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa$

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

2)

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

$p=\frac{F}{A}$

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

$p=56.1\cdot 10^6 Pa$ is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

$A=4\pi r^2$

Therefore, we can find the force exerted on it by re-arranging the previous equation:

$F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N$

6 0

2 years ago