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worty [1.4K]
2 years ago
9

Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in 8 s. Scenario C: 200 J of work is done in 10 s. W

hich scenario uses the most power?
Physics
2 answers:
hodyreva [135]2 years ago
8 0
Using the equation P = W/t to solve your problem . 

Thus the answer is all of them use the same amount of power. 20 J.  
Mice21 [21]2 years ago
6 0

Answer:

They all used the same power.

Explanation:

Power is equal to the ratio between the work done (W) and the time taken (t):

P=\frac{W}{t}

Let's calculate the power used in the three scenarios:

A) P=\frac{120 J}{6 s}=20 W

B) P=\frac{160 J}{8 s}=20 W

C) tex]P=\frac{200 J}{10 s}=20 W[/tex]

So, the power used is the same in the three scenarios.

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Simora [160]
The unit 'mb' means millibar which is equivalent to 1/1000 of 1 bar. To convert the units from bar to atmospheres (atm) and to inches Hg (inHg), we need to know the conversion factors.

a.) 1 atm = 1.01325 bar

0.92 mb(1 bar/1000 mbar)(1 atm/1.01325 bar) =<em> 9.08×10⁻⁴ atm</em>

b.) 1 bar = 29.53 inHg

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3 0
2 years ago
You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond
luda_lava [24]

Answer:

a) L = 0.75m   f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m   f₁ = 56.67 Hz ,  f₃ = 170 Hz

Explanation:

This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung

In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1  node         λ₁ = 4 L

With 2 nodes      λ₂ = 4L / 3

With 3 nodes       λ₃ = 4L / 5

The general term would be      λ_n= 4L / n         n = 1, 3, 5, ((2n + 1)

The speed of sound is

         v = λ f

         f = v / λ

         f = v  n / 4L

Let's consider each length independently

L = 0.75 m

        f₁ = 340 1/4 0.75 = 113.33 n

         f₁ = 113.33 Hz

        f₃ = 113.33   3

       f₃ = 340 Hz

L = 1.5 m

       f₁ = 340 n / 4 1.5 = 56.67 n

       f₁ = 56.67 Hz

       f₃ = 56.67 3

       f₃ = 170 Hz

8 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
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agasfer [191]

Answer:

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5 0
2 years ago
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