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Alex787 [66]
3 years ago
12

Describe the changes in the field of view and the amount of available light when going from low to high power using the compound

microscope
Physics
1 answer:
kogti [31]3 years ago
8 0
A light microscope is a common lab tool used to study microscopic organisms and structures. When increasing the magnification (power) of the microscope, the field of view is decreased, since the microscope is essentially "zooming" in to a smaller area. For a given numerical aperture, the amount of available light decreases when going to a higher magnification (M) because the light collection is proportional to 1 / M^2.
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How often should cardio exercises be performed? A. Never B. Three to five times a month C. Three to five times a week D. Three t
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C. Three to five time a week.

Explanation:

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A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge
zheka24 [161]

Answer:

The acceleration of an electron is 1.2\times10^{20}\ m/s^2

Explanation:

Given that,

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Distance on y axis = 3.0 cm

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Distance on x axis = 4.0 cm

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Using formula of force

F_{1}=\dfrac{kq_{1}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}

F_{1}=8\times10^{-11}j\ \ N

We need to calculate the force on electron due to q₂

Using formula of force

F_{2}=\dfrac{kq_{2}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}

F_{2}=6.93\times10^{-11}i\ \ N

We need to calculate the net force

Using formula of net force

F=F_{1}+F_{2}

Put the value into the formula

F=8\times10^{-11}j+6.93\times10^{-11}i

The magnitude of the net force

F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}

F=1.058\times10^{-10}\ N

We need to calculate the acceleration of an electron

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F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}

a=1.2\times10^{20}\ m/s^2

Hence, The acceleration of an electron is 1.2\times10^{20}\ m/s^2

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Answer:

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