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larisa86 [58]
3 years ago
14

If you wanted to learn more about what it was like to work in a family of scientists, whose life would you research?

Physics
2 answers:
atroni [7]3 years ago
8 0

Answer:

marie

Explanation:

Schach [20]3 years ago
6 0

Answer:

Stephen hawking if his family were scientists

You might be interested in
5. A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5th minimum occurs at 7.00° aw
abruzzese [7]

For a single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen,the angle  is mathematically given as

theta=25.3

Option A is correct

<h3> What angle does the 18th minimum occur?</h3>

Generally, the equation for the the angle   is mathematically given as

\theta=n(\lambda/d)

Therefore

\theta 1/ \theta 2=n1(\lambda/d)/ n2(\lambda/d)

In conclusion

theta/7=16/5

theta=10*7/5

theta=25.3

Read more about Angle

brainly.com/question/14362353

3 0
2 years ago
The formula for speed is v = s/t. <br> a. True<br> b. False
crimeas [40]
If you mean S is the distance then it is true 
Velocity = Distance / time 
3 0
3 years ago
An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete
Ksju [112]
 <span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:


R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km 

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s 

  the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h 

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu!  :)</span>
6 0
3 years ago
What does it mean for water to have a high heat capacity
Aliun [14]

Answer:

Its heat capacity is higher than that of any other liquid or solid, its specific heat being 1 cal / g, this means that to raise the temperature of 1 g of water by 1 ° C it is necessary to provide an amount of heat equal to a calorie . Therefore, the heat capacity of 1 g of water is equal to 1 cal / K.

Explanation:

The water has a very high heat capacity, a large amount of heat is necessary to raise its temperature 1.0 ° K. For biological systems this is very important because the cellular temperature is modified very little in response to metabolism. In the same way, aquatic organisms, if water did not possess that quality, would be very affected or would not exist.

This means that a body of water can absorb or release large amounts of heat, with little temperature change, which has a great influence on the weather (large bodies of water in the oceans take longer to heat and cool than the ground land). Its latent heats of vaporization and fusion (540 and 80 cal / g, respectively) are also exceptionally high.

6 0
3 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
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