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spayn [35]
3 years ago
13

Two small insulating spheres with radius 9.00*10^-2m are separated by a large center-to-center distance of 0.520m . One sphere i

s negatively charged, with net charge -2.40uC , and the other sphere is positively charged, with net charge 3.35uC . The charge is uniformly distributed within the volume of each sphere.
a)What is the magnitude E of the electric field midway between the spheres?
Take the permittivity of free space to be ?0 = 8.85
Physics
1 answer:
ankoles [38]3 years ago
4 0

Answer:

electric field  7.64 10⁵ N / C

Explanation:

For this problem we can use Gauss's law, which indicates that the load within a Gaussian surface can be considered located in its center, therefore, as the point is much greater than the radius of the spheres we can take them as if it were a spot loading Let's calculate the field at the midpoint between them

            r = 0.520 m / 2

            r = 0.260 m

The electric field is a vector

            Et = E1 + E2

     

E1 has a negative charge, so the force is attractive and E2 has a positive charge, the outside is repulsive, so the two forces point in the same direction on the positive test charge.

               E1 = k Q1 / r²

               E2 = k Q2 / r²

               Et = E1 + E2

               Et = k Q1 / r² + k Q2 / r² = k (Q1 + Q2) / r²

               Et = 8.99 10⁹ (2.4 10⁻⁶ + 3.35 10⁻⁶) / 0.26²

               Et = 7.64 10⁵ N / C

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wrong statement :  Momentum is not conserved for a system of objects in a head-on collision.

Explanation:

In a head on collision of two objects , two equal and opposite forces are created at the point of collision . These two forces create two impulses in opposite direction which results in equal and opposite changes in momentum in each of them . Hence net change in momentum is zero. In this way momentum is conserved in head on collision of two objects.

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<span>radiation, hydrogen, and helium </span>
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How can you determine the amount of work done on an object? (joules, work, power...etc.)
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a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde
Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

  • The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal
  • The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

Volume \ of \ a \ cone = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot h

  • Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

  • \dfrac{dV}{dr} = \dfrac{d}{dr}  \left(\dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot  s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

\underline{\left  \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}

\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}

Learn more here:

brainly.com/question/14466080

3 0
3 years ago
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
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Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
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