PLEASE HELPPP

2 answers:

8 0

This is what I found online: "<span>For a person with normal hearing, when it comes to pitch the human hearing range starts low at about 20 Hz. That’s about the same as the lowest pedal on a pipe organ. On the other side of the human hearing range, the highest possible frequency heard without discomfort is 20,000Hz. While 20 to 20,000Hz forms the absolute borders of the human hearing range, </span>our hearing is most sensitive in the 2000 - 5000 Hz frequency range<span>." So i'm guessing it's 100,00 since none of those are lower than 20

</span>

agasfer [191]3 years ago

4 0

The answer is 100,000 hz

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How to change V=72km/hr to m/s

lapo4ka [179]

Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).

Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.

(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)

= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)

= 20 meter/second

7 0

4 years ago

A motorist drives north for 36.0 minutes at 96.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km

Over [174]

Answer:

a) $d=187.6km$

b) $v=61.5km/h$

Explanation:

First we convert our minutes to hours so we work always in the same units.

$36min=36min(\frac{1h}{60min})=0.6h$

$15min=15min(\frac{1h}{60min})=0.25h$

Where we used the fact that 1 hour are 60 min, thus the multiplying factor is equal to 1 (not altering the time, just changing the units).

a) On the first part the motorist travels a distance $d_1=v_1t_1=(96km/h)(0.6h)=57.6km$ , and on the second part he travels $d_2=130km$ .

The total displacement is $d=d_1+d_2=57.6km+130km=187.6km$

b) The average velocity is the relation between the total displacement and the time taken to cover it. Our total time is t=0.6h+0.25h+2.2h=3.05h, thus we have:

$v=\frac{187.6km}{3.05h}=61.5km/h$

8 0

3 years ago

A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo

Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$