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3 years ago

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Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

tric field 4 cm from the plate is:

1 answer:

AysviL [449]3 years ago

7 0

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

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JulsSmile [24]

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

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4 years ago

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Your brother is insisting that you’ll never hear a sound produced behind a barrier wall at the end of your yard you notice that

Tresset [83]

Answer

D.Diffraction

Explanation

Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.

The ways tends to curve behind the barrier. This is called diffraction of waves.

Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction

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3 years ago

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ly charged particles are held 24 x 103m apart and then released from rest. The initial acceleration of the first particle is obs

inessss [21]

Answer:

Part a)

$m_2 = 4.9 \times 10^7 kg$

Part b)

$q_1 = q_2 = 5312.6 C$

Explanation:

Part a)

As we know that both charge particles will exert equal and opposite force on each other

so here the force on both the charges will be equal in magnitude

so we will have

$F = m_1a_1 = m_2a_2$

here we have

$6.3 \times 10^7(7) = m_2(9)$

now we have

$m_2 = 4.9 \times 10^7 kg$

Part b)

Now for the force between two charges we can say

$F = \frac{kq_1q_2}{r^2}$

now we have

$(6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}$

now we have

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3 years ago

A jet airliner moving initially at 548 mph

sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

$v_1 =548 mph$ along the x-direction

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To find the resultant, we must resolve both vectors on the x- and y- axis:

$v_{1x}= 548 mph$

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3 years ago

If light is used to promote the electron, what are the selection rules for the initial and final spin and orbital angular moment

loris [4]

Answer:

(1) $\Delta S=0$

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Explanation:

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The conditions or selection rule to promoting the electron are discussed below:

(1) The total spin should not change that is $\Delta S=0$ .

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3 years ago