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Katen [24]
3 years ago
13

Who has the most power? athlete A (bench presses 100 kg over 0.6 m in 0.5 s) athlete B (bench presses 150 kg over 0.6 m in 1.0 s

) athlete C (bench presses 200 kg over 0.6 m in 2.0 s) athlete D (bench presses 250 kg over 0.6 m in 2.5 s)
Physics
1 answer:
MrRissso [65]3 years ago
4 0

Answer:

Athlete A

Explanation:

Power is the rate of doing work and it is calculated as follows:

Power = work done/time taken = mgh/t

(for work being done against gravity)

So for athlete A

P = (100 kg * 9.8 N/kg* 0.6m)/0.5 s = 1176 W

For athlete B

P = (150 kg * 9.8 N/kg* 0.6m)/1 s = 882 W

For athlete C

P = (200 kg * 9.8 N/kg* 0.6m)/2 s = 588 W

For athlete D

P = (250 kg * 9.8 N/kg* 0.6m)/2.5 s = 588

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3 years ago
Explain why atoms only emit certain wavelengths of light when they are excited. Check all that apply. Check all that apply. Elec
JulsSmile [24]

Answer:

Explanation:

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>FALSE. </em>The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.

The energies of atoms are not quantized. <em>FALSE. </em>The energies of the atoms are in specific levels.

When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. <em>FALSE. </em>During absorption, a specific wavelength of light is absorbed, not emmited.

Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>TRUE. </em>Again, you can observe just the transition due the change of energy of an electron in the quantized energy level

When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. <em>TRUE. </em>The electron decreases its energy releasing a specific wavelength of light.

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7 0
3 years ago
IBM has a fast computer that it calls the Blue Gene/L that can do '136.8
dmitriy555 [2]

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

6 0
3 years ago
Which statement below is false? A) Weight depends on the force of gravity.
Scilla [17]

I think it's the letter Did (this has to be 20 characters long) it would be Different or would be D

5 0
3 years ago
Read 2 more answers
electricity flows back and forth in an lc circuit with a .33 f capacitor and a .1 h inductor. what is the angular frequency and
Shalnov [3]

The time it takes an object to complete one oscillation and return to its initial position is measured in terms of a period, or T. The formula for the angular frequency is = 2/T.

<h3>How is G determined in oscillation?</h3>

Use a stopwatch to calculate the oscillation's time period T. Calculate the pendulum's length L. Subtract the time period T's square from the length L.

<h3>How does oscillation's G work?</h3>

A mass attached to the end of a pendulum with a length of l causes it to oscillate with a period (T). T = 2(l/g), where g.

To know more about angular frequency visit:-

brainly.com/question/29107224

#SPJ4

4 0
1 year ago
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