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Irina-Kira [14]

3 years ago

12

Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart

. What is the electric field (in N/C) at the center of the region between the plates? (Enter the magnitude.)

1 answer:

Romashka [77]3 years ago

5 0

Answer:

E = 5291.00 N/C

Explanation:

Expression for capacitance is

$C = \frac{\epsilon A}{d}$

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

$C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}$

$C = 24.06 \epsilon$

$C = 24.06\times 8.854\times 10^{-12} F$

$C =2.1\times 10^{-10} F$

We know that capacitrnce and charge is related as

$V = \frac{Q}{C}$

$= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}$

v = 9.523 V

Electric field is given as

$E = \frac{V}{d}$

= $\frac{9.52}{1.8*10^{-3}}$

E = 5291.00 V/m

E = 5291.00 N/C

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Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

3 years ago

In normal light conditions, how well do plants grow when there are 10 plants in the

Sati [7]

Answer

when there are ten they don't grow so well but when there is less than 10 they tend to grow

7 0

3 years ago

James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s

liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{James}=\sqrt{\frac{2h}{g}}$

Vertical motion of John : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{John}=\sqrt{\frac{2h}{g}}$

So both times are same.

Both of them reach the lake at the same time.

3 0

3 years ago

I need help with these. Please show workings<br>

Sauron [17]

Answer:

Imp = 25 [kg*m/s]

v₂= 20 [m/s]

Explanation:

In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.

1)

$(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})$

where:

m₁ = mass of the object = 5 [kg]

v₁ = initial velocity = 0 (initially at rest)

F = force = 5 [N]

t = time = 5 [s]

v₂ = velocity after the momentum [m/s]

$(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]$

2)

$(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]$

8 0

2 years ago

Read 2 more answers

solve the following system by any method 2x - 6y = 24 ... -5x + 6y = -6 ... A. (0,-6) B. (4,-1) C. (-6,-6) D. (6,-1)

vodka [1.7K]

Using elimination, the answer is C.

5 0

3 years ago