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frozen [14]
3 years ago
15

I love making friends but i always feel lonely around them, i try to act fine and as if nothing bothers me but it does. I want a

steady partner who loves me for me. Is it to much to ask for? People are mad because I'm clingy which is something I'm working on but I don't like people hating me for little things I do. I try to make friends but everyone makes it hard for me. I'm sorry I made you read this.. but sometimes talking about these things help..
Physics
2 answers:
arsen [322]3 years ago
6 0

Answer:

i will be your friend ...

Ivan3 years ago
4 0

Answer:

Explanation:

its good you let it out, just remember youre not alone. just keep trying to improve on the things you want to improve on and focus on yourself. if you feel lonely around the friends you made, then they might not be good for you, i know its hard to let go because they are the only people you have but trust me, its better if you let go of them and take time to really get to know yourself before getting to know anyone else.

before you get a steady partner, learn to love yourself. it sound like you didnt  but are trying now which is great! keep it going because before you can love ANYBODY you must love yourself. Keep fighting and i'll be rooting for you!!

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Answer:

The answer is charcoal

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Which of the following is characteristics of both sound and light waves
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==> they carry energy from one place to another;

==> their speed depends on properties of the medium;

==> they undergo reflection, refraction, diffraction, and dispersion;

==> their intensity varies as the inverse square of the distance they travel;

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3 years ago
How is the voltage (electrical potential energy) of the current affected as it passes through a resistor?
Margaret [11]
Electrical voltage is the potential energy of an electrical supply stored in the form of an electrical charge. From the equation I = V/R, the current flowing through a circuit is directly proportional to the voltage and inversely proportional to the resistance.Therefore, the voltage of the current increases as it passes through a resistor. 
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3 years ago
A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla
Fiesta28 [93]

Answer:

a) P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}=  0.349 W

b) For this case we se that P \propto A^2 f^2

Since the power is constant but the frequency is doubled we will see that A^2 \propto \frac{P}{f^2}

So the original amplitude is A_i \propto \sqrt{\frac{P}{f^2}}

And if the frequency is doubled we have that:

A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}

A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}

So then we will see that the amplitude would be reduced the half and for this case:

A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m

Explanation:

For this case we have the following data given:

D= 1mm = 0.001 m represent the diameter

r = D/2= 0.0005m represent the radius

T= 5.7 N represent the tension

f = 57 Hz represent the frequency of the oscillator

A= 0.54 cm = 0.0054 m represent the amplitude of the wave

Part a

For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:

P= 2 \pi^2 \rho S v f^3 A^2

This expression can be written in different ways:

P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2

P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2

P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}

Where f is the frequency , T the tension rho= 7800 \frac{kg}{m^3} the density of the steel, A the amplitude and S= \pi r^2 the area, so then we have everuthing in order to replace and we got:

P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}=  0.349 W

Part b

For this case we se that P \propto A^2 f^2

Since the power is constant but the frequency is doubled we will see that A^2 \propto \frac{P}{f^2}

So the original amplitude is A_i \propto \sqrt{\frac{P}{f^2}}

And if the frequency is doubled we have that:

A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}

A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}

So then we will see that the amplitude would be reduced the half and for this case:

A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m

8 0
3 years ago
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<span>Resolution = wavelength / ((2) (numerical aperture))
Resolution = 500 nm / (2 ) ( 1.25) = 200 nm  = 0.2 um</span>
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