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lyudmila [28]
3 years ago
10

An infinitely large plane carries a uniformly distributed charge such that the charge per unit area is 3.47E-6 C/m2. What is the

electric field magnitude |E| at a point that is 2.40 m from the plane
Physics
1 answer:
kodGreya [7K]3 years ago
3 0

Answer:1.95\times 10^5\ N/C

Explanation:

Given

The charge per unit area on the plate is \sigma =3.47\times 10^{-6}\ C/m^2

Electric for a infinitely large plane is given by

E=\dfrac{\sigma }{2\epsilon_o}

Therefore, electric field at a point 2.4 m from the plane is

\Rightarrow E=\dfrac{3.47\times 10^{-6}}{2\times 8.854\times 10^{-12}}\\\\\Rightarrow E=0.195\times 10^6\ N/C

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A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the ro
irina1246 [14]

Answer:

2.45 J

Explanation:

The following data were obtained from the question:

Mass (m) = 0.5 kg

Height (h) = 1 m

Kinetic energy (KE) =?

Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 1/2 = 0.5 m

Final velocity (v) =?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 0.5)

v² = 9.8

Take the square root of both side

v = √9.8

v = 3.13 m/s

Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:

Mass (m) = 0.5 kg

Velocity (v) = 3.13 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 3.13²

KE = 0.25 × 9.8

KE = 2.45 J

Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J

8 0
3 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Mars2501 [29]

Answer:

Therefore % increase in velocity is 18.23 %

Explanation:

we use the equality of mass flow rate and the areas

m_1 = m_2\\p_1v_1 = p_2v_2\\p_1A_1v_1 = p_2A_2v_2\\v_2 = \frac{p_1}{p_2} v_1

The percentage increase in velocity is

Δ v% = \frac{v_2 - v_1}{v_1} \\100%

= \frac{p_1}{p_2} v_1 - v_1.100%

= \frac{\frac{1.2}{1.015} - 1}{1} . 100%

= Therefore % increase in velocity is 18.23 %

5 0
3 years ago
Following are the different layers of the sun’s atmosphere. Rank them based on the order in which a probe would encounter them w
blsea [12.9K]

Answer:

Going from earth to the sun a probe would encounter the next layers in order:

  • Corona
  • Transition Region
  • Chromosphere
  • Photosphere
  • Convection Zone
  • Radiative Zone
  • Core

A brief description of them:

Corona is the outermost layer and it cannot  be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.

Transition Region is between the corona and the chromosphere, it has an extension of about 100km

The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.

The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.

The convection zone is where convection happens, hot gas rises, cools and rises again.

Radiative Zone is where the photons try to rise to move to higher layers.

The core of the Sun is where nuclear fusion occurs due to the very high temperatures.

6 0
3 years ago
If you double the wavelength, the electromagnetic radiation energy will double. The energy of the electromagnetic radiation will
Sedbober [7]

Answer:

It's energy will double.

Explanation:

This is because energy, E, is related to frequency, f, by:

E = hf

Where h = Planck's constant

So, double frequency will be 2f

=> E(2f) = 2hf = 2E.

Hence, energy is doubled.

7 0
3 years ago
A 2.00 kg object is moving in a circular path with a radius of 5.00 cm. The object starts from rest and with constant angular ac
Rom4ik [11]

Answer:

0.800 m/s²

Explanation:

First, calculate the angular acceleration:

ω = αt + ω₀

6.00 rad/s = α (3.00 s) + 0 rad/s

α = 2.00 rad/s²

Now calculate the angular velocity at t = 2.00 s:

ω = αt + ω₀

ω = (2.00 rad/s²) (2.00 s) + 0 rad/s

ω = 4.00 rad/s

Calculate the linear velocity:

v = ωr

v = (4.00 rad/s) (0.0500 m)

v = 0.200 m/s

Finally, calculate the centripetal acceleration:

a = v² / r

a = (0.200 m/s)² / (0.0500 m)

a = 0.800 m/s²

3 0
3 years ago
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