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Zielflug [23.3K]
2 years ago
14

Jose has a mass of 70 kg, what is his weight​

Physics
1 answer:
rusak2 [61]2 years ago
6 0
His weight would be 154.32
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You are making a round trip from City A to City B and back to City A again at constant speed. At what point in the trip is your
MA_775_DIABLO [31]

Answer:

Halfway between B and A on the return leg.

Explanation:

Your average SPEED for the entire trip will equal your constant speed as the time and distance increase at proportionate rates.

Your average VELOCITY will equal your constant speed while you travel from A to B because time and displacement are increasing at proportionate rates.

When you turn around at B to return, your Displacement is now decreasing while your travel time continues to increase, so your average velocity decreases.

Lets say the distance from A to B is 90 km and your constant speed is 30 km/hr.

your average speed is 30 km/hr because you took 6 hrs to travel 180 km

We want to find your position when your average velocity is 30/3 = 10 km/hr

it took 3 hrs to go 90 km from A to B. Let t be the time lapsed since turn around

your displacement is given by d = 90 - 30(t)

and your total time of travel is t + 3 hrs

 v = d/t

10 = (90 - 30t) / (t + 3)

10(t + 3) = (90 - 30t)

10t + 30 = 90 - 30t

40t = 60

t = 1.5 hrs

This will occur when you are halfway between B and A

3 0
2 years ago
A. How will the Milky Way continue to move outward as a result of the Big Bang?
emmasim [6.3K]

Answer:

i believe it is c

Explanation:

4 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
A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of
GREYUIT [131]

Given:

Sample 1:

Chloroform is CHCl_{3}

12 g Carbon

1.01 g Hydrogen

106.4 g Cl

Sample 2:

30.0 g of Carbon

Solution:

mass of chloroform from sample 1:

12 + 1.01 +106.4 =119.41 g

Now, for the total mass of chloroform in sample 2:

mass of chloroform \times\frac{given mass of carbon}{mass of carbon atom}

mass of chloroform = 119.41 \times\frac{30}{12} = 298.53 g

6 0
3 years ago
The ______ is the component of communication through which information is transferred.
MrMuchimi
The medium is the component of communication through which information is transferred. The correct option in the given question is option "c". Medium is actually the means through which communication is done. It can be a an oral message or written message or it can be nonverbal or symbolic. Any kind of message transmission requires a medium and without an medium no message can be transmitted. Whenever people talk on the telephone or write a letter, there has to be a medium for transmitting the message to the person on the other side.
3 0
2 years ago
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