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Shtirlitz [24]
3 years ago
15

an electron is very far away from a proton (fixed in place) and is released from rest. how fast will the electron be travelling

when it is 0.03 m away from the proton?
Physics
1 answer:
lions [1.4K]3 years ago
3 0
The electron won't be traveling because it is as u say "fixed in place".
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A volleyball is dropped from a cliff and a soccer ball is thrown upward from the same position. When each ball reaches the groun
vredina [299]
The question is whether the statement is true or false.

The answer if false.

Explanation:

It is exactly the opposite. The soccer ball will hit the ground with greater velocity.

Since the soccer ball is thrown upward, when it returns to the same heigth from which it was throwm it will have a velocity downward, which will make that the soocer ball reaches the ground at the bottom of the clif with greater velocity than the volleball.

The greater the velocity with which the soccer ball is thrown upward, the greater its velocity when reaches the same point from which it was thrown, and the greater the velocity with which it will hit the ground at the bottom of the clif.
6 0
3 years ago
Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. Th
xxTIMURxx [149]

The amount of charge that passes per unit time is called <em>electric current</em> .

Current has dimensions of [Charge] / [Time] .

It's measured and described in units of ' Ampere ' .

1 Ampere means 1 Coulomb of charge passing a point every second.

7 0
3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so
lubasha [3.4K]

Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

4 0
4 years ago
Light of wavelength 480 nm illuminates a pair of slits separated by 0.27 mm. If a screen is place 1.7 m from the slits, determin
Dahasolnce [82]

Answer:

  Δy = 6.05 mm

Explanation:

The double slit phenomenon is described by the expression

      d sin θ = m λ                constructive interference

      d sin θ = (m + ½) λ       destructive interference

      m = 0,±1, ±2, ...

As they tell us that they measure the dark stripes, we are in a case of destructive interference, let's use trigonometry to find the sins tea

      tan θ = y / x

      y = x tan θ

In the interference experiments the measured angle is very small so we can approximate the tangent

      tan θ = sin θ / cos θ

     cos θ = 1

     tan θ = sin θ

     y = x sin θ

We substitute in the destructive interference equation

     d (y / x) = (m + ½) λ

    y = (m + ½) λ x / d

The first dark strip occurs for m = 0 and the third dark strip for m = 2. Let's find the distance for these and subtract it

 m = 0

      y₀ = (0+ ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

      y₀ = 1.511 10⁻³ m

 m = 2

     y₂ = (2 + ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

     y₂ = 7.556 10⁻³ m

The separation between these strips is Δy

     Δy = y₂-y₀

     Δy = (7.556 - 1.511) 10⁻³

     Δy = 6.045 10⁻³ m

     Δy = 6.05 mm

5 0
3 years ago
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