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SashulF [63]
3 years ago
5

An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.35 μC/m2. Another infini

te sheet of charge with uniform charge density σ2 = -0.54 μC/m2 is located at x = c = 31 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 13.5 cm and x = 17.5 cm).. 1)What is Ex(P), the x-component of the electric field at point P, located at (x,y) = (6.75 cm, 0)?. 2). What is σa, the charge density on the surface of the conducting slab at x = 13.5 cm?
Physics
1 answer:
Maslowich3 years ago
3 0
Using a cylinder for a gaussian surface we find the flux through each end means that the surface integral will be double (the field has to be split between both directions, if you like)<span>∮<span>E⃗ </span>⋅<span>d⃗ </span>A=E(2A)=<span>q<span>ϵ0</span></span>=<span><span>σA</span><span>ϵ0</span></span>→E=<span>σ<span>2<span>ϵ0</span></span></span></span>which does not depend on distance from the slab. The conducting slab in the middle picks up an induced charge from each charged slab, so it does not contribute to or hinder the field strength. Hence the total of the E-fields at any point should be<span><span>∑E=<span>E1</span>+<span>E2</span>=<span><span>σ1</span><span>2<span>ϵ0</span></span></span>+<span><span>σ2</span><span>2<span>ϵ0</span></span></span></span></span>
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From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s
Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

8 0
3 years ago
A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh
kolezko [41]

Answer:

The angular velocity is  w = 53.35 \ rounds /minute

Explanation:

From the question we are told that

    The mass of the block is  m = 61.2kg

     The of the pulley is  M = 14.2 kg

      The radius of the pulley is  R = 1.5m

       The radius  of the cord around the pulley is  r = 1.5 m

       The distance of the block to the floor is  d = 8.0 m

         

From the question we are told that the moment of inertia of the pulley is

          I  = \frac{1}{2} MR^2 kg \cdot m^2

Substituting value  

         I = \frac{1}{2}  * 14.2 * (1.5)^2

         I = 15.975 kg \cdot m^2

Using the Newtons law we can express the force acting on the vertical axis as

              ma = mg -T

         =>  T = mg -ma

Now when the pulley is rotated that  torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

                  \tau = I \alpha

     Here \alpha is the angular acceleration

           Here \tau is the torque which can be equivalent to

              \tau = T r

  Substituting this above

            Tr = I \alpha      

Substituting for T

         (mg - ma ) r =  I\  r \alpha

Here a is the  linear acceleration which is mathematically represented as

           a = r\alpha

    (mg - m(r\alpha ) ) r =  I\  r \alpha

     mgr = I\alpha  + m(r\alpha ) r

    mgr = \alpha  [ I + mr^2]

   making \alpha the subject

          \alpha  = \frac{mgr}{I -mr ^2}          

   Substituting values

            \alpha  = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}

             \alpha =5.854 rad /s^2

Now substituting into the equation above to obtain the acceleration

             a = 5.854 * 1.5

                a=8.78 m/s^2

This acceleration is a = \frac{v}{t}

and v is the linear velocity with is mathematically represented as

         v = \frac{d}{t}

Substituting this into the formula acceleration

        a = \frac{d}{t^2}

making t the subject

         t = \sqrt{\frac{d}{a} }

substituting value

      t = \sqrt{\frac{8}{8.78}}

     t = 0.9545 \ s

Now the linear velocity is

       v = \frac{8}{0.9545}

       v = 8.38 m/s

The angular velocity is  

       w = \frac{v}{r}

So

       w = \frac{8.38}{1.5}

        w = 5.59 rad/s

Generally 1 radian is equal to  0.159155 rounds or turns

        So  5.59 radian is  equal to x

Now x is mathematically obtained as

         x = \frac{5.59 * 0.159155}{1}

            = 0.8892 \ rounds

 Also

      60  second =  1 minute

So   1 second  = z      

Now z is mathematically obtained as

         z = \frac{ 1}{60}

            z = 0.01667 \ minute

Therefore

              w = \frac{0.8892}{0.01667}

              w = 53.35 \ rounds /minute

           

8 0
3 years ago
Danny lowers the sails on his boat. He paddles upstream at 19 km/hr. The current is still running downstream at 15 km/hr. What i
Molodets [167]

Answer:

4 km/hr

Explanation:

The computation of the actual velocity is shown below:

Because the path of its paddles is opposed to the current direction, the real velocity can be determined by deducting the current velocity to its velocity while paddling

So, the actual velocity is

= Upstream - downstream

= 19 km/hr - 15 km/hr

= 4 km/hr

As we can see it is in positive, so it is an upstream direction  

3 0
3 years ago
A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

For more information on this visit

brainly.com/question/21811998

4 0
2 years ago
Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half
lutik1710 [3]
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
Therefore, the average velocity is 
v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h
0 0
3 years ago
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