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Svetllana [295]
3 years ago
15

What are the magnitude and direction of electric field at the center of square?

Physics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

hello the diagram related to this question is missing attached below is the missing diagram

Answer :

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

Explanation:

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

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If 31.25
valentina_108 [34]
<span>1 C = 6.24150965(16)×10^18 electrons

31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs

</span><span>I hope this helps. </span>
4 0
3 years ago
Read 2 more answers
Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball
Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

              y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e)    y = 0 + ½ g t²

     t = √ (2y / g)

     t = √(2 125 / 9.8)

     t = 5.05 s

6 0
3 years ago
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various
Schach [20]

Answer:

q  =  -461532.5 \ C

Explanation:

From the question we are told that

     The  electric filed is  E  =  102 \ N/C  

Generally according to Gauss law

=>   E  A  =  \frac{q}{\epsilon_o }

Given that  the electric field is pointing downward  , the equation become

    - E  A  =  \frac{q}{\epsilon_o }

Here   q is the excess charge on the surface of the earth

          A is the surface  area of the of the earth which is mathematically represented as

     A  =  4\pi r^2

Where r is the radius of the earth which has a value r = 6.3781*10^6 m

 substituting values

    A  = 4 * 3.142  *   (6.3781*10^6 \ m)^2

    A  =5.1128 *10^{14} \ m^2

So

   q  =  -E  * A  *  \epsilon _o

Here \epsilon_o s the permitivity of free space with value

          \epsilon_o  =  8.85*10^{-12} \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

     q  =  -102  * 5.1128 *10^{14}  *  8.85 *10^{-12}

     q  =  -461532.5 \ C

6 0
4 years ago
Write the equation of motion gover the centre of mass<br><br><br>​
Nataly [62]

Answer:

write the equation of motion go over the centre of mass

Explanation:

the center of mass of a distribution of mass in space (sometimes referred to as the balance point) is the unique point where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.

6 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
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