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Nikolay [14]
3 years ago
8

Two charges +q and -q are situated at certain distance, at the point exactly midway betwee them what happens on electric field a

nd potential?
Physics
1 answer:
Luden [163]3 years ago
7 0

Answer: Electric field is not zero but potential is zero

Explanation:

In science, charge, also recognized as electric charge, electrical charge, or electrostatic charge and expressed q, is a component of a unit of body that reveals the extent to which it has more or fewer electrons than protons.

Considering the direction of the electric field is of positive to negative charge. Therefore, it determination be toward due −q also due to both the charges.

But for potential, it is a scalar quantity, so the sum of potential due to all the charges will be zero.

Henceforth, the option (c) Electric field is not zero but potential is zero is the correct answer.

a incorect

From the above statements, we could conclude that option (c) Electric field is not zero, but potential is zero is the correct answer. So, the option (a) Electric field and potential both are zero is an incorrect solution.

b incorect

Among all the options, the option (c) Electric field is not zero but potential is zero is the correct answer.

Therefore, the option (b) Electric field is zero but potential is not zero is an incorrect solution

d) is incorrect

It has been explained that the option (c) Electric field is not zero but potential is zero is the correct answer.

Hence, the option (d) Electric field is not zero but potential is zero is an incorrect solution.

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How fast is a cat that runs 50 meters in 10 seconds?
saw5 [17]

Answer:

50 m in 10 sec….divide 50 by 10 to convert to metres per second = 5 m/s. Then multiply by 3.6 (3600 seconds in an hour divided by 1000) to convert to kilometres per hour =18 m/s. To convert to mph divide by 1.609 =11.18 mph.

Explanation:

5 0
3 years ago
Read 2 more answers
IE is the energy required to remove an electron from an atom. As atomic radius increases, the valence electrons get farther from
Bond [772]

Answer:

The bigger the atom the lesser the ability of the atom to hold on to its valence electrons.

Explanation:

Atomic radius can be looked at as the distance between the nucleus and the outermost energy level. As an atom gets bigger, the outer shell gets further and further from the positive nucleus. this means that electrons that are in the outer energy level become less held (attracted) by the nucleus because of distance and shielding of the attractive forces by the electrons in the lower energy levels. This means that as an atom becomes bigger, its ability to hold on to its outer electrons lessens.

4 0
3 years ago
The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresp
IRISSAK [1]

Answer:

a

   k    =  11600000 N/m

b

   \Delta  L  =  3.2323 *10^{-5} \ m

c

  F =  3750.28 \  N  

Explanation:

From the question we are told that

    The Young modulus is  E =  1.4 *10^{10} \  N/m^2

     The length is  L  =  0.35 \ m

      The  area is  2.9 \ cm^2  =  2.9 *10^{-4} \ m ^2

   

Generally the force acting on the tibia is mathematically represented as

       F =  \frac{E *  A  *  \Delta  L }{L}    derived from young modulus equation

Now this force can also be mathematically represented as

      F =  k *  \Delta  L    

So

     k    =  \frac{E *  A  }{L}

substituting values

     k    =  \frac{1.4 *10^{10} *  2.9 *10^{-4}  }{ 0.35}

     k    =  11600000 N/m

    Since the tibia support half the weight then the force experienced by the tibia is  

        F_k  =  \frac{750 }{2}  =  375 \  N

 From the above equation the extension (compression) is mathematically represented as

          \Delta  L  =  \frac{ F_k  *  L  }{ A *  E }        

substituting values

           \Delta  L  =  \frac{  375   *  0.35  }{ (2.9 *10^{-4}) *   1.4*10^{10} }

           \Delta  L  =  3.2323 *10^{-5} \ m

From the above equation the maximum force is  

        F =  \frac{1.4*10^{10} *  (2.9*10^{-4})  *  3.233*10^{-5} }{ 0.35}  

         F =  3750.28 \  N  

4 0
3 years ago
Two blocks, joined by a string, have masses of 6.0 and 9.0 kg. They rest on a frictionless, horizontal surface. A second string,
Lynna [10]

Answer:

T= 27 N

Explanation:

Assuming that the string joining both masses is massless  and inextensible, both masses accelerate at the same rate.

So, we can treat to both masses as a single system, and apply Newton's 2nd Law to both masses.

In this way, we can get the value of the acceleration without taking into account the tension in the string, as it is an internal force (actually a action-reaction pair).

Newton's 2nd law is a vector equation, so we can decompose the forces along perpendicular axis in order to convert it in two algebraic equations.

We can choose one axis as parallel to the horizontal surface (we call it x-axis, being the positive direction the one of  the movement of the blocks due to the horizontal force applied to the 6.0 kg block), and the other, perpendicular to it, so it is vertical (we call y-axis, being the upward direction the positive one).

Taking into account the forces acting  on both masses, we can write both equations as follows:

Fy = N- (m₁+m₂)*g = 0 (as there is no movement in the vertical direction)

Fx = Fh = (m₁ + m₂) * a ⇒ 45 N = 15.0 kg * a

⇒ a = 45 N / 15.0 kg = 3 m/s²

Now, in order to get the value of the tension T, we can choose as our system, to any mass, and apply Newton's 2nd Law again.

If we choose to the mass of 6.0 kg, in the horizontal direction, there are two forces acting on it, in opposite directions: the  horizontal applied force of 45 N, and the tension in the string that join both masses.

The difference of both forces, must be equal to the mass (of this block only) times the acceleration, as follows:

F- T = m₂* a ⇒ 45 N - T = 6.0 kg * 3 m/s²

⇒ T = 45 N -18 N = 27 N

We could have arrived to the same result taking the 9.0 Kg as our system, as the only force acting in the horizontal direction is just the tension in the string that we are trying to find out, as follows:

F = m₁*a = 9.0 kg* 3 m/s² = 27 N

4 0
3 years ago
an object moving at 10. km/hr has a kinetic energy of 10. J. what is the kinetic energy of the same object if it is moving at 20
Schach [20]
Kinetic energy is related to velocity by:
KE = (1/2)mv^2

solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
10= 50m
10/50 = m
1/5 = m

at 20 km/hr

KE = (1/2)(1/5)(20)^2
KE = (1/10)(400)
KE = 40 J
5 0
3 years ago
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