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tatuchka [14]
3 years ago
13

Light travels in a straight line at a constant speed of 300000 m/s. What is the acceleration of light?

Physics
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

As light travels in a straight line at a constant speed, it's acceleration is <u>0 m/s²</u>.

There is no rate of change of speed, so there is no acceleration.

  • <u>0 m/s²</u> is the right answer.
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An ideal gas is confined within a closed cylinder at atmospheric pressure (1.013 * 105 Pa) by a piston. The piston moves until t
likoan [24]

Answer:

911700\ \text{Pa}

Explanation:

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V_2= Final volume = \dfrac{V_1}{9}\\\Rightarrow \dfrac{V_1}{V_2}=9

Temperature is the same in the initial and final state

From the ideal gas law we have

P_1V_1=P_2V_2\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=P_1\times9\\\Rightarrow P_2=1.013\times 10^5\times 9\\\Rightarrow P_2=911700\ \text{Pa}

The final pressure of the system is 911700\ \text{Pa}.

5 0
3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch
vladimir1956 [14]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is 3.2\cdot 10^7 N/C (radially inward)

C) The electric field at 6.00 cm from the surface is 1.2\cdot 10^7 N/C (radially inward)

Explanation:

A)

We can solve the problem by applying Gauss Law, which states  that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

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

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

\epsilon_0 is the vacuum permittivity

By taking a sphere centered in the origin,

\int E dS = E \cdot 4\pi r^2

where 4\pi r^2 is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

R=9.0 cm = 0.09 m (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

q=0

And therefore,

E4\pi r^2 = 0\\\rightarrow E = 0

So, the electric field inside the plastic sphere is zero.

B)

Here we apply again Gauss Law:

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

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

r=R=0.18 m

The charge contained within the Gaussian sphere is therefore

q=-29.0 \mu C = -29.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

E=3.2\cdot 10^7 N/C

C)

For this part again, we apply Gauss Law:

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

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

q=-29.0 \mu C = -29.0\cdot 10^{-6}C

Therefore, the electric field in this case is

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C

And again, this is radially inward, so according to the sign convention asked in the problem,

E=1.2\cdot 10^7 N/C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0
3 years ago
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