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Rama09 [41]
3 years ago
15

A. It shows the relative distance of each planet from the sun

Physics
1 answer:
Scilla [17]3 years ago
3 0

Answer:

C. It shows the relative size of each planet tone another

Explanation:

I took this quiz earlier. Hope you have a great day! :)

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Why are nuclear power plants controversial
VashaNatasha [74]
I would have to say because while they do produce relatively clean power, it is extremely dangerous and overly harmful should one "melt down." Some people believe the reward is worth the risk, others do not. It is more a matter of perspective - are you near the nuclear power plant or not? If you are, chances are you won't like it - on the other hand, if you are not right hear the power plant you might like it more. I hope this helps!

P.S. - If you think I did a good job I would love it if you went to my YouTube channel to see if I can help you more there. I am just starting out and don't have much time to record, but I would love to have some user-submitted topics! Here is the link: https://www.youtube.com/channel/UCG0zQcSsxZ2mu-OnZRdI2-Q. I hope to see you there!
8 0
3 years ago
Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At s
Tresset [83]

Answer:

n=0.03928 moles

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

Explanation:

The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs

Volume of oxygen in lungs =V=5*20%= 1 L=1*10^{-3} m^3

Temperature=T=37^oC=273+37=310K

Pressure at sea level = P= 1 atm=1.0125*10^5 Pa

R is universal Gas Constant =8.314 J/mol.K

Formula:

n=\frac{PV}{RT}\\n=\frac{(1.0125*10^5) *(1*10^{-3})}{(8.314)*310} \\n=0.03928 mol

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

8 0
4 years ago
Light from an argon laser strikes a diffraction grating that has 4,917 lines per cm. The first-order principal maxima are separa
miss Akunina [59]

Answer:

Wavelength is 4.8x10^-7m

Explanation:

See attached file

3 0
3 years ago
The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally
dalvyx [7]
(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
K= \frac{1}{2}mv^2
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 2.20 \cdot 10^{-15} J}{9.1 \cdot 10^{-31} kg} }=6.95 \cdot 10^7 m/s

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
qvB = m \frac{v^2}{r}
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
r= \frac{mv}{qB}= \frac{(9.1 \cdot 10^{-31} kg)(6.95 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(3.00 \cdot 10^{-5} T)}=13.18 m

And finally we can calculate the centripetal acceleration, given by:
a_c =  \frac{v^2}{r}= \frac{(6.95 \cdot 10^7 m/s)^2}{13.18 m}=3.66 \cdot 10^{14} m/s^2
5 0
3 years ago
Waves travel along a 100-m length of string which has a mass of 55 g and is held taut with a tension of 75 N. What is the speed
Dvinal [7]

Answer:

v=369.27\frac{m}{s}

Explanation:

The speed of the waves in a string is related with the tension and mass per unit length of the string, as follows:

v=\sqrt\frac{T}{\mu}

First, we calculate the mass per unit length:

\mu=\frac{m}{L}\\\mu=\frac{55*10^{-3}kg}{100m}\\\mu=5.5*10^{-4}\frac{kg}{m}

Now, we calculate the speed of the waves:

v=\sqrt\frac{75N}{5.5*10^{-4}\frac{kg}{m}}\\v=369.27\frac{m}{s}

8 0
3 years ago
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