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malfutka [58]
3 years ago
11

HELP ASAP PLS. ILL GIVE BRAINLIEST

Physics
1 answer:
Blizzard [7]3 years ago
4 0

Answer:

I'm not completely sure, but I believe the first and third of the three are mechanical.

Explanation:

Chemical potential isn't moving or about to go into motion. It can't be mechanical.

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Muscles use force to move objects. When using a ramp to do work, would your muscles use more force or less force?
White raven [17]
Less because the ramp is letting off force but i does depend on the way you are going on the ramp

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3 years ago
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How do you tell a stars life expectancy based on an hr diagram?
e-lub [12.9K]
The X-axis of the H-R Diagram indicates the star's surface temperature in degrees Kelvin.  The Y-axis, on the other hand, indicates luminosity, or brightness.

Main sequence refers to a roughly diagonal, slightly S-curved line stretching between the upper-left and lower-right corners on which main sequence stars chart. They maintain a predictable relationship between luminosity and temperature: the brighter, the hotter. The upper-right quadrant of the H-R diagram is home to newly discovered red giants while the lower-left quadrant of the H-R Diagram belongs almost exclusively to white dwarfs.
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3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
3 years ago
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an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o
ZanzabumX [31]

Answer:

The maximum amplitude (V_{max}) will be 7.96 V.

Explanation:

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S = 2 \pi f V_{max}

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Therefore, from the above equation we can write,

&&5 \frac{V}{\mu s} = 2 \pi 100 kHz \times V_{max}\\&or,& V_{max} = \frac{5V}{10^{-6} s \times 2 \pi 100 \times 10^{3} Hz}\\&or,& V_{max} = \frac{5}{2 \pi \times 10^{-1}} V = 7.96 V

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5 0
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