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Fittoniya [83]
1 year ago
12

Calculate the magnitude of the electric field at the location of given that the square is 6.05 cm on a side.

Physics
1 answer:
Serggg [28]1 year ago
4 0

The term “electric field” refers to the physical field that surrounds electrically charged particles and acts to either attract or repel all other charged particles in the field (also known as an E-field).

It can also refer to the physical field surrounding a system of charged particles. Electric fields are composed of electric charges and time-varying electric currents.

Both electric and magnetic fields are manifestations of the electromagnetic field, one of the four fundamental interactions (sometimes known as forces) of nature.

Electrical technology makes use of electric fields, which are significant in many branches of physics.

For instance, in atomic physics and chemistry, the electric field acts as an attractive force to hold atoms’ atomic nuclei and electrons together. It is also the force that causes atoms to chemically link together to form molecules.

To know more about electric field visit:

brainly.com/question/14811118

#SPJ4

You might be interested in
Which statement is the correct representation of these electric field lines?
GuDViN [60]
I would say a. hope this helps
4 0
3 years ago
Read 2 more answers
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo
gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

                       H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

 Height at any time 'T' after the drop =

                          (initial height) +

                                              (initial velocity) x (T) +
                                                                 (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

     Upward  = the positive direction

                       Initial height = +150 m
                       Initial velocity = + 3 m/s

     Downward = the negative direction

                     Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

                 H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
0    =  (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

                   -4.9 T²  +  3T  + 150  =  0

Use the quadratic equation:

                         T  =  (-1/9.8) [  -3 plus or minus √(9 + 2940)  ]

                             =  (-1/9.8) [  -3  plus or minus  54.305  ]

                             =  (-1/9.8) [ 51.305  or  -57.305 ]

                          T  =  -5.235 seconds    or    5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

                    H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
H'(t)  =  v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set                                  v₀ + a T  =  0

                                      +3  -  9.8 T  =  0

Add 9.8 to each  side:   3               =  9.8 T

Divide each side by  9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh !  I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
                       (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) =  1.5 m/s .

Distance climbed during that time = (average speed) x (time)

                                                           =  (1.5 m/s) x (0.306 sec)

                                                           =  0.459 meter  (hardly any at all)

     But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at  150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely  5.847 seconds after its release.  

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work.  The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0
3 years ago
This is an object's change in motion per unit time in a specified direction.
timurjin [86]

Answer:

Velocity

Explanation:

Velocity is an object's change in motion per unit time in a specified direction

7 0
3 years ago
Read 2 more answers
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