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3 years ago

I've got a test tomorrow morning help!

Physics

1 answer:

topjm [15]3 years ago

3 0

Answer:

The efficiency is just 0.016

Explanation:

The efficiency is given by the useful energy ÷ by the total energy.

1 W is the same that 1 joule per second

If you do the math:

1.6/100=0.016

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A rock dropped on the moon will increase its speed from 0m/s to 8.15 m/s in 5 seconds what is the acceleration of the rock

sergejj [24]

speed Δv = v₂ = 8.15ms⁻¹

v₁ = 0ms⁻¹

time =t = 5s

acceleration = a = speed / time taken

a = Δv / Δt ( as Δv = v₂ - v₁)

a = 8.15ms⁻¹ / 5 s

a = 1.6ms⁻²

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4 years ago

A bus travels 54km in 90 minutes.calculte the speed of the bus

storchak [24]

Speed = distance/time
Time = 90/60 = 1.5 hours
Speed = 54/1.5
Speed = 36km/h

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3 years ago

Which of these statements would best explain the problem encountered with nuclear waste disposal? A) The isotopes have a long ha

algol13

That's right, the correct answer is
<span>A) The isotopes have a long half-life and only remain radioactive for a long time period

The half life of an isotope is the time it takes for the amount of the sample to reduce to half of its initial value. If an isotope has a long half-life, it means it takes a long time to reduce down to a significant level, so it will remain radioactive for a long time period.</span>

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4 years ago

Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not

d1i1m1o1n [39]

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

$I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},$

where $I_0(\text{at }1000\text{ Hz})$ is the hearing threshold at 1000 Hz.

Therefore we have the following

$0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}$

$I_1$ is the threshold at 1000Hz and $I_2$ is the threshold at 100Hz.

By exponentiating we have

$10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.$

Now dividing these two equations we get

$\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.$

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.

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3 years ago

A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will

Hitman42 [59]

Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.

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3 years ago