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charle [14.2K]
2 years ago
11

the diagram below shows the situation described in the problem. the focal length of the lens is labeled f; the scale on the opti

cal axis is in centimeters. draw the three special rays, ray 1, ray 2, and ray 3, as described in the tactics box above, and label each ray accordingly. do not draw the refracted rays.
Physics
1 answer:
Artist 52 [7]2 years ago
8 0

Answer:

I have no clue

Explanation:

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A small 16 kilogram canoe is floating downriver at a speed of 4 m/s. What is the canoe's kinetic energy?
seraphim [82]

Kinetic Energy,K.E=1/2MV²

mass,m=16kg

velocity,v=4m/s

K.E=1/2×16×4²

=128kgm²/s²

=128 Joules

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3 years ago
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Why are Watson’s experiments considered controversial?
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D, it is considered unethical today
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A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha
ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, L=400\ \mu H=400\times 10^{-6}\ H

Capacitance of parallel LCR circuit, C=0.3\ \mu F=0.3\times 10^{-6}\ F

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

f=\dfrac{1}{2\pi \sqrt{LC} }

f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }

f=14528.79\ Hz

or

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

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3 years ago
What is the relationship between frequency ad wavelenght
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Answer:

Frequency and wavelength are inversely proportional to each other. The wave with the greatest frequency has the shortest wavelength. Twice the frequency means one-half the wavelength. For this reason, the wavelength ratio is the inverse of the frequency ratio.

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A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

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