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Bezzdna [24]
3 years ago
7

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

l to zero at this point? If not, then provide an example to prove your answer. If you think that the answer is "yes", please justify it.
Physics
1 answer:
lyudmila [28]3 years ago
8 0

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

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A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor
VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596 kg-m2

I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2

by perpendicular axis theorem

I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2

4 0
3 years ago
Two forces are acting on an object, but the net force on the object is O N. For the net
Alenkinab [10]

Answer:

  A.  The forces are the same size and in opposite directions.

Explanation:

Just as an opposite number will cancel a number: -1 +1 = 0, so an opposite force will cancel a force, with the result that the net is zero.

6 0
3 years ago
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A vector has a size and a direction. The size is called the magnitude of the vector.
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What height would cause an object to have no gravitational potential energy
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The height is zero.
That means whatever place is your reference for height measurement,
the object is lying right there at that height.
5 0
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