Answer: To isolate the unknown substances, of which only tiny amounts were present, the Curies were the first to use a new method of chemical analysis. They employed various standard (but sometimes demanding) chemical procedures to separate the different substances in pitchblende.
Explanation:
Answer:
C. 590 mph
![\vert v_{cj}\vert=589.49\ mph](https://tex.z-dn.net/?f=%5Cvert%20v_%7Bcj%7D%5Cvert%3D589.49%5C%20mph)
Explanation:
Given:
- velocity of jet,
![v_j=500\ mph](https://tex.z-dn.net/?f=v_j%3D500%5C%20mph)
- direction of velocity of jet, east relative to the ground
- velocity of Cessna,
![v_c=150\ mph](https://tex.z-dn.net/?f=v_c%3D150%5C%20mph)
- direction of velocity of Cessna, 60° north of west
Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.
Refer the attached schematic.
So,
![\vec v_j=500\ \hat i\ mph](https://tex.z-dn.net/?f=%5Cvec%20v_j%3D500%5C%20%5Chat%20i%5C%20mph)
&
![\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)](https://tex.z-dn.net/?f=%5Cvec%20v_c%3D150%5Ctimes%20%28%5Ccos120%5C%20%5Chat%20i%2B%5Csin120%5C%20%5Chat%20j%29)
![\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph](https://tex.z-dn.net/?f=%5Cvec%20v_c%3D-75%5C%20%5Chat%20i%2B75%5Csqrt%7B3%7D%5C%20%5Chat%20j%5C%20mph)
Now the vector of relative velocity of Cessna with respect to jet:
![\vec v_{cj}=\vec v_j-\vec v_c](https://tex.z-dn.net/?f=%5Cvec%20v_%7Bcj%7D%3D%5Cvec%20v_j-%5Cvec%20v_c)
![\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )](https://tex.z-dn.net/?f=%5Cvec%20v_%7Bcj%7D%3D500%5C%20%5Chat%20i-%28-75%5C%20%5Chat%20i%2B75%5Csqrt%7B3%7D%5C%20%5Chat%20j%20%29)
![\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph](https://tex.z-dn.net/?f=%5Cvec%20v_%7Bcj%7D%3D575%5C%20%5Chat%20i-75%5Csqrt%7B3%7D%5C%20%5Chat%20j%5C%20mph)
Now the magnitude of this velocity:
![\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}](https://tex.z-dn.net/?f=%5Cvert%20v_%7Bcj%7D%5Cvert%3D%5Csqrt%7B%28575%29%5E2%2B%2875%5Csqrt%7B3%7D%20%29%5E2%7D)
is the relative velocity of Cessna with respect to the jet.
The lowest energy of electron in an infinite well is 1.2*10^-33J.
To find the answer, we have to know more about the infinite well.
<h3>What is the lowest energy of electron in an infinite well?</h3>
- It is given that, the infinite well having a width of 0.050 mm.
- We have the expression for energy of electron in an infinite well as,
![E_n=\frac{n^2h^2}{8mL^2}](https://tex.z-dn.net/?f=E_n%3D%5Cfrac%7Bn%5E2h%5E2%7D%7B8mL%5E2%7D)
![m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1](https://tex.z-dn.net/?f=m%3D9.1%2A10%5E%7B-31%7Dkg%5C%5CL%3D0.050%2A10%5E%7B-3%7Dm%5C%5Ch%3D6.63%2A10%5E%7B-34%7DJs%5C%5Cn%3D1)
- Thus, the lowest energy of electron in an infinite well is,
![E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J](https://tex.z-dn.net/?f=E_1%3D%5Cfrac%7B%286.63%2A10%5E%7B-34%7D%29%5E2%7D%7B8%2A9.1%2A10%5E%7B-31%7D%2A%280.050%2A10%5E%7B-3%7D%29%7D%3D1.2%2A10%5E%7B-33%7DJ)
Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.
Learn more about the infinite well here:
brainly.com/question/20317353
#SPJ4
Answer:
20 seconds.
Explanation:
The following data were obtained from the question:
Distance = 10 m
Speed = 0.5 m/s
Time =...?
The speed of an object is simply defined as the distance travelled by the object per unit time. Mathematically, it is expressed as:
Speed = Distance /time
With the above formula, we can obtain the time taken for the ball to travel a distance of 10 m as shown below:
Distance = 10 m
Speed = 0.5 m/s
Time =...?
Speed = Distance /time
0.5 = 10/time
Cross multiply
0.5 × time = 10
Divide both side by 0.5
Time = 10/0.5
Time = 20 secs.
Therefore, it will take 20 seconds for the ball to travel a distance of 10 m.
Answer:
Rate of change of magnetic flux
Explanation:
The induced current is equal to the ratio of induced emf to the resistance of the conductor.
According to the Faraday's law of electromagnetic induction, the induced emf is proportional to the rate of change of magnetic flux.