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lesantik [10]
2 years ago
10

The plates of a parallel-plate capacitor are oppositely charged and attract each other. Find the expression for the force one pl

ate exerts on the other.
Physics
1 answer:
Mamont248 [21]2 years ago
5 0

The force exerted by one plate for two parallel plates arrangement is F = QE/2.

<h3>Force one plate exerts on the other</h3>

The force exerted by one plate for two parallel plates arrangement is given as follows;

F = QE/2

where;

  • Q is the charge on one plates
  • E is the electric field due to the charge
  • F is force exerted by one plate

Thus, the force exerted by one plate for two parallel plates arrangement is F = QE/2.

Learn more about force between parallel plates here: brainly.com/question/13590045

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SOMEBODY PLEASE HELP!!! Indicate the reasons why the centripetal acceleration (and centripetal force) always point to the center
maxonik [38]
Well we know the correct answer cannot be "a" bcause velocity is tangent to the circlular path of an object experienting centripical motion. Velocity DOES NOT point inward in centripical motion.

we know the correct answer cannot be "b" because "t" stands for "time" which cannot point in any direction. so, time cannot point toward the center of a circle and therefore this answer must be incorrect.

I would choose answer choice "c" because both force and centripical acceleration point toward the center of the circle.

I do not think answer choice "d" can be correct because the velocity of the mass moves tangent to the circle. velocity = (change in position) / time. Therefore, by definition the mass is moving in the direction of the velocity which does not point to the center of the circle.

does this make sense? any questions?

3 0
3 years ago
A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal.
Korvikt [17]

Answer:

B) 27.3 m

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x =  vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 30 m/s , at an angle  α=40.0° above the horizontal

v₀x = vx = 30*cos40° = 22.98 m/s

v₀y = 30*sin40° = 19.28 m/s

y₀ = 2m

y =  18.0 m

g = 9.8 m/s²

Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

(vfy)² = 58.26

v_{f} = \sqrt{58.26}

vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

Horizontal distance from where the rock was thrown to the window

We replace t = 1.19 s , in the equation (1)

x =  vx*t  

x = (22.98)* ( 1.19 )

x = 27.3 m

3 0
3 years ago
what are the common elements that make up series and parallel circuits , and how are circuits diagrammed?​
Ahat [919]
<h2>Answer:</h2>

A series circuit occurs when the elements are connected along a simple path so the same current flows through all the elements. On the other hand, a parallel circuit occurs when there are two or more paths for the electricity to flow. The diagram are shown in the Figure below. We have chosen a source and resistors to illustrate this problem.

3 0
4 years ago
A ball is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
vovangra [49]

Answer:

a) 48.5 ft/s

b) 36.5 ft

c) -80.3 ft/s

Explanation:

a)

The equation of motion of the ball is :

y(t) = -16.1 ft/s^2 * t^2 + Vo*t

Where Vo is the initial velocity

If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

Vo  = (16.1*25- 160) ft / 5s = 48.5 ft/s

b)

To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)

y(1.5s) = 36.52 ft

c)

We now need the time at which y(t') = -64 ft

-64 = -16.1*t'^2 + 48.5*t'

By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s

3 0
3 years ago
If a car moving down this road rounds this curve and changes its direction to the right, is this an example of speed, velocity,
mars1129 [50]

Answer: velocity

Explanation: it's the rate of change of the objects position/ consistent change

4 0
3 years ago
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