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vladimir1956 [14]
3 years ago
10

5. A 1.2 m guitar string is under a tension of 888 N. The waves produced on the string (when

Physics
1 answer:
Usimov [2.4K]3 years ago
4 0
I don’t even know tbh
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If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.
elena-14-01-66 [18.8K]
Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
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Please Help!! When do ionic bonds occur?
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3 years ago
What is the strength of the electric field in a region where the electric potential is constant?
Roman55 [17]

Answer:

Where the electric potential is constant, the strength of the electric field is zero.

Explanation:

As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e

Eₓ = - dV / dx        ----------(i)

From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.

<em>Therefore, a constant electric potential means that electric field is zero.</em>

4 0
3 years ago
What is 0 point energy times negative 0 point energy ????????
elixir [45]
1 point energy should be the answer
8 0
3 years ago
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po
zvonat [6]

Answer:

0.293I_0

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

I_1=\frac{I_0}{2}

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

I_2=I_1 cos^2 \theta

where

\theta is the angle between the axes of the two polarizer

Here we have

\theta=40^{\circ}

So the intensity after the 2nd polarizer is

I_2=I_1 (cos 40^{\circ})^2=0.587I_1

And substituting the expression for I1, we find:

I_2=0.587 (\frac{I_0}{2})=0.293I_0

5 0
3 years ago
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