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2 years ago

10

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

electric field between the two plates?

1 answer:

Stells [14]2 years ago

8 0

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

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Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

$W = q \Delta V$

where

$q=3.0 \mu C = 3.0 \cdot 10^{-6} C$ is the magnitude of the charge

$\Delta V = 6.0 kV = 6000 V$ is the potential difference between point P and infinity

Substituting into the equation, we find

$W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J$

4 0

2 years ago

Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw

Natasha_Volkova [10]

Answer:

Explanation:

Given

average speed of Phelps $v_{avg}=2\ m/s$

total distance $d=200\ m$

he swims first 100 m at an average speed if $1.8 m/s$

so time taken is $t_1=\frac{100}{1.8}=55.55\ s$

suppose $t_2$ is the time taken to swim remaining half

average velocity is $v_{avg}=\frac{displacement}{total\ time}$

$v_{avg}=\frac{100+100}{55.55+t_2}$

$t_2+55.55=\frac{200}{2}=100$

$t_2=44.45\ s$

so velocity in the second half is

$v_2=\frac{100}{45.45}$

$v_2=2.19\approx 2.2\ m/s$

3 0

2 years ago

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor

nalin [4]

Answer:

$u_x=38.13\ m/s$

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

$S=u_yt+\dfrac{1}{2}gt^2$

Here $u_y=0$

S= 57 cm

$0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2$

t=0.34 s

Now in the horizontal direction

$x=u_xt$

Here x=13 m

t= 0.34 s

So

$13=u_x\times 0.34$

$u_x=38.13\ m/s$

So the initial speed of ball is 38.13 m/s.

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1 year ago

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

madreJ [45]

In order to answer this exercise you need to use the formulas

S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s

7 0

2 years ago

Read 2 more answers

Given a triangle with sides X = 6.35 cm and Y = 12.25 cm with an angle of 90 degrees between them, find the length of the hypote

never [62]

Answer:

the hypotenuse = 13.78 cm

Ф = 27.44°

θ = 62.56°

explanation:

4 0

2 years ago