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3 years ago

What are the 3 major uses of water? Describe each in 3-4 sentence

Physics

1 answer:

kirza4 [7]3 years ago

5 0

Answer:

-Drinking and Household Needs.

-Recreation.

-Industry and Commerce.

-Agriculture.

-Thermoelectricity/Energy.

Explanation:

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A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.0 amu) makes up 75.8% of the sample, and t

irinina [24]

Answer:

$M_{av}=35.521amu$

Explanation:

As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

$M_{av}=(35amu*75.8+37amu*24.3)/100=35.521amu$

4 0

3 years ago

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

$\omega=\frac{v}{r}$

Then, you solve the equation (1) in order to obtain v/r:

$\frac{v}{r}=\omega=\frac{qB}{m}$

Finally, you replace the values of the parameters:

$\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}$

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

$f=2\pi \omega=5.5*10^7Hz$

5 0

3 years ago

The breaking in (blank) bonds in food releases energy for your body to use

kicyunya [14]

Answer:

The breaking in <em>molecular</em> bonds in food releases energy for your body to use.

5 0

2 years ago

A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

5 0

3 years ago

What does the diameter in meters need to be of a round parachute that the designer has determined needs to have an area of 500 s

Marrrta [24]

Area of a circle is
A= pi r^2
so
500m^2 = 3.14 r2
500/pi = r ^2
152.1549...=r^2
square root both sides
r=12.61566...
d=2r
d=25.2
to 3 sig fig

4 0

3 years ago