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steposvetlana [31]
3 years ago
10

18. A swimmer steps out of a swimming pool into a light breeze. Why does

Physics
1 answer:
OverLord2011 [107]3 years ago
6 0
When the swimmer gets out, the water on her skin begins to evaporate. This process requires takes a lot of energy and that energy comes from the heat of her body. This is similar to when someone sweats. Hope this helps!
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You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-
Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

E_P=\int dE \cos

E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0
3 years ago
An electron moves along the z-axis with vz=4.5×10^7m/s. As it passes the origin, what are the strength and direction of the magn
Hitman42 [59]
This answer of the strength and directions of the magnetic field is a
5 0
3 years ago
Which of the following is not in the form of waves?
Nady [450]
C. Seismic energy
This is energy that is released in earthquakes.
6 0
3 years ago
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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq
gayaneshka [121]

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law,  PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}

Given;

\frac{dP}{dt} (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) =  600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( \frac{dv}{dt});

(600 x 40) + (150 x \frac{dv}{dt}) = 0

\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min

Therefore, the volume is decreasing at 160 cm³/min

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3 years ago
Why are yet planes faster tharn. cargo planes?
stellarik [79]

Answer:

cargo planes hold cargo so there hevier

Explanation:

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