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3 years ago

Why is there no electric field at the center of a charged spherical conductor?

1 answer:

4 0

-- Since the sphere is a conductor, the charge on it will move around
until it's evenly distributed on the surface of the sphere. When every
tiny smitch of charge is the same distance from the charge around it,
they won't need to move around any more.

-- At that point, the situation at the center of the sphere will be:
For every smitch of charge on the surface, causing en electric field
at the center, there is another smitch, of the same size and in exactly
the opposite direction, canceling out the field of the first one.
Every smitch of charge on the surface causes a tiny bit of electric field
at the center, and they all cancel each other.

It turn out that if the sphere is hollow, then the electric field at EVERY
point inside it is zero, not only at the center.

It's exactly the same idea as a sphere with uniform, homogeneous mass.
For that sphere, the gravity at the center is zero.

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How does inertia affect a person who is not wearing a seatbelt during a collision?

son4ous [18]

When the car stops, inertia causes the person to continue forward at the speed of the car prior to stopping

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3 years ago

Read 2 more answers

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adelina 88 [10]

Answer:

Explanation:

7 0

3 years ago

How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and

Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{9}{40}$

$f=4.44\ cm$

When , R₁= -4, R₂ = 8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{3}{40}$

$f=-13.33\ cm$

When , R₁= 4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{3}{40}$

$f=13.33\ cm$

When , R₁= -4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{9}{40}$

$f=-4.44\ cm$

Hence, The lenses with different focal length are four.

8 0

3 years ago

A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?

Alex17521 [72]

<h2>Answer:53.63 $ms^{-2}$ </h2>

Explanation:

The equations of motion used in this question is $v=u+at$

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of $9.8ms^{-2}$ .

In X-Direction,

Given that initial velocity= $u_{x}$ = $33.8ms^{-1}$

Using $v=u+at$ ,

$v_{x}=33.8+(0)4.25=33.8ms^{-1}$

In Y-Direction,

Given that initial velocity= $u_{x}$ = $0ms^{-1}$

Using $v=u+at$ ,

$v_{y}=0+(9.8)4.25=41.65ms^{-1}$

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}$

7 0

3 years ago

What is the purpose of using significant figures? How does it relate to accuracy, precision, resolution, and uncertainty?

Umnica [9.8K]

Answer:

#see solution for details

Explanation:

-Uncertainty refers to an estimate of the amount by which a result may differ from this value,

-Precision refers to how closely repeated measurements agree with each other.

-Accuracy refers to how closely a measured value agrees with the correct value.

-The number of significant figures is the number of digits believed to be correct by the person doing the measuring. Therefore, choosing the correct number of significant figures reduces the deviation from the point of accuracy/uncertainty or precision and thereby reducing margin of error in the ensuing calculations.

7 0

3 years ago