What are the best units for DENSITY of a solid like a rock or a lemon? *

A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T

Mazyrski [523]

Given a capacitor that is charge by a battery

So when the battery was disconnect, the charge remains on the capacitor

One plate will have positive charge

And the other will have a negative charge but of same magnitude

Now, the plates are pulled apart, so the distance between them was extended

a. The capacitance of the capacitor?

The capacitance of a capacitor is given as

C = εo•A/d

Where C is the capacitance

εo is permissivity, a constant

A is are of plate

d is the distance apart

From the formula we notice that

Capacitance is inversely proportional to distance, so this implies that as the distance is increase, the capacitance is reduce.

Capacitance Decreases

b. Potential difference?

The potential difference is given as

q = CV

V = q/C

Or V=Ed

Where E is electric field

So the potential difference is inversely proportional to the capacitance, so we know that the capacitance is reducing, then voltage will increase.

Or, using the second relations

V=Ed, voltage is directly proportional to distance, So, if the distance apart is increasing then, the voltage is increasing

Voltage is increasing

c. Electric field between plate.

The electric field is given as

E = kQ/r²

So the electric field is inversely proportional to the distance square, so as the distance is an increase, the electric field is decreased.

Electric field decreases

d. Electric potential energy?

U = ½CV², then, q = CV

Then, U = ½qV

The electric potential energy is directly proportional to the voltage and since the voltage is increasing due to increase in distance, then the electric potential energy increases.

3 0

3 years ago

Someone help me pls

TiliK225 [7]

Explanation:

The dog was stationary at segment c

8 0

3 years ago

Read 2 more answers

A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra

seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial speed of the projectile

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

$\theta=60^{\circ}$ (angle)

Solving the equation, we find the horizontal range:

$d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m$

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity of the bullet after having covered a vertical displacement of $s$

$u_y$ is the initial vertical velocity

$a =-g=$ is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

$u_y = u sin\theta$

Also, the maximum height s is reached when the vertical velocity becomes zero,

$v_y =0$

Substituting into the equation and re-arranging for s, we find the maximum height:

$s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

3 0

3 years ago

What is one disadvantage of a series circuit?

nlexa [21]

If one bulb goes out then all the others won't light up because electricity will be cut off. It's a disadvantage because in a parallel circuit if one bulb burns out all the others will still be on because they won't be affected. I hope I've helped you ☺

6 0

3 years ago