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xz_007 [3.2K]
3 years ago
12

Hey i am quite confused on this question, if someone could help me understand it i would really appreciate! :)

Physics
2 answers:
Alekssandra [29.7K]3 years ago
8 0

You have to create table using cubic meters and the unit Kelvin to measure the temperature.

yarga [219]3 years ago
7 0

I did an experiment. Experiment was to strech tungsten until it fractured. My "tensometer" had a manual graph plotter. So I was able to plot the force vs extension of the tungsten.

When it broke, there was a loud bang. And it was hot to touch.

qualitative abive.

-------

quantitaive data could have been had I measured ad recorded the temperature of

the tungsten, and recorded the force and extension in a table of values.

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) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha
vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

f= 466 Hz

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s

The angular frequency instead is given by

\omega = 2\pi f

And substituting

f = 466 Hz

We find

\omega = 2\pi (466 Hz)=2926.5 rad/s

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz

While the angular frequency is given by:

\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s

(c) 4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s

The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s

(d) 2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s

In this case, the frequency is

f=5.0 MHz = 5.0 \cdot 10^6 Hz

So the period in this case is

T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

While the angular frequency is given by

\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s

7 0
3 years ago
I need help please somebody help me
Morgarella [4.7K]

Answer:

Generally, magnets are attracted to objects that are made of the metals iron, nickel, or cobalt. These materials are called ferromagnetic materials. ... When all or most of the domains are aligned in the same direction, the whole object becomes magnetized in that direction and becomes a magnet.

Explanation:

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2 years ago
What determines how soon one action potential can follow another?
belka [17]
It is through biopsychological feedback. 

A class of chemical called a neurotransmitter is important in the transmission of nerve impulses. Neurotransmitters are packaged by the cell into small, membrane-bound sacs called vesicles. Upon receiving a chemical signal, the vesicles move toward the cell membrane and fuse with it, releasing the enclosed neurotransmitters from the terminal end of the nerve cell. 
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Which is correct? I need to know!!
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What website are you using?
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3 years ago
After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m>s. Her 5.0-kg cat
Vinil7 [7]

To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_1= mass of the skier

m_2= mass of the cat

v_1 = initial velocity of skier

v_2 = initial velocity of cat

v_f= final velocity of both

Re-arrange to find V_f we have,

V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}

V_f = \frac{(60)(15)+(5)(-3.8)}{(60+5)}

V_f = 13.55m/s

Once the final velocity is found it is possible to calculate the change in kinetic energy, so

\Delta KE = KE_i-KE_f

\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2

\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2

\Delta KE = 819.1J

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0
3 years ago
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