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3 years ago

An object moves along the x axis according to the equation x = 3.00t^2 - 2.00t + 3.00 where x is in meters and t is in seconds.

2 answers:

8 0

   x = 3 t² - 2 t + 3

When t=2 sec, x = 3 (4) - 2 (2) + 3
   = 12    - 4     + 3 = 11 meters .

When t=3 sec,    x = 3 (9) - 2 (3) + 3
   = 27 - 6    + 3 = 24 meters .

Distance = (end position) - (start position) = (24m - 11m) = 13 meters

Time = (end time) - (start time) = (3s - 2s) = 1 second

Speed = (distance) / (time) = 13 meters / 1 sec = 13 meters/second

MakcuM [25]3 years ago

7 0

We are given the equation:

x = 3 t² - 2 t + 3
We use that equation to calculate for the distance traveled.

At t=2 sec, x = 3 (4) - 2 (2) + 3
x =11 meters

At t=3 sec, x = 3 (9) - 2 (3) + 3
x = 24 meters

Velocity = (24 meters - <span>11 meters) / (3 s - 2 s ) = 13 m/s</span>

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By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
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Now we can use the following equation to calculate the length of the wire:
$R= \frac{\rho L}{A}$ (1)
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In this problem, we have a wire of copper, with resistivity $\rho=1.68 \cdot 10^{-8} \Omega m$ . The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m$
And the cross-sectional area is
$A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2$

So now we can rearrange eq.(1) to calculate the length of the wire:
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4 years ago

A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari

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Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency c/λ where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

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