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faltersainse [42]
4 years ago
14

How far does a wave travel in three cycles?

Physics
1 answer:
eduard4 years ago
6 0
It will be 3 wavelengths because 1 cycle = 1 wavelength. 
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The reflection of a sound wave from a wal is called
WINSTONCH [101]
I think its an echo?
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A student is given a sample of matter and asked to determine whether it is an element.
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I think the correct answer is C
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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 large truck has a mass of 20000kg. it is traveling at 28m/s along a staright road . calculate the kinetic energy ​
Semmy [17]

Answer:

The answer is

<h2>7,840,000 J</h2>

Explanation:

The kinetic energy KE of an object given it's mass and velocity can be found by using the formula

KE =  \frac{1}{2} m {v}^{2}  \\

where

m is the mass

v is the velocity

From the question

m = 20000kg

v = 28 m/s

It's kinetic energy is

KE =  \frac{1}{2}  \times 20000 \times  {28}^{2}  \\  = 10000 \times 784

We have the final answer as

<h3>7,840,000 J</h3>

Hope this helps you

3 0
3 years ago
Problem 3.2 Part A Draw the vector C⃗ =A⃗ +B⃗ .(Figure 1)
Elodia [21]
I can't answer this question without a figure. I've found a similar problem as shown in the first picture attached. When adding vectors, you don't have to add the magnitudes only, because vectors also have to factor in the directions. To find the resultant vector C, connect the end tails of the individual vectors. <em>The red line (second picture) represents the vector C.</em>

5 0
3 years ago
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