Answer:
The y-component of the electric force on this charge is ![F_y = -1.144\times 10^{-6}\ N.](https://tex.z-dn.net/?f=F_y%20%3D%20-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20N.)
Explanation:
<u>Given:</u>
- Electric field in the region,
![\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%28148.0%5C%20%5Chat%20i-110.0%5C%20%5Chat%20j%29%5C%20N%2FC.)
- Charge placed into the region,
![q = 10.4\ nC = 10.4\times 10^{-9}\ C.](https://tex.z-dn.net/?f=q%20%3D%2010.4%5C%20nC%20%3D%2010.4%5Ctimes%2010%5E%7B-9%7D%5C%20C.)
where,
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
![\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%5Cdfrac%7B%5Cvec%20F%7D%7Bq%7D%5C%5C%5Ctherefore%20%5Cvec%20F%20%3D%20q%5Cvec%20E%5C%5C%3D%2810.4%5Ctimes%2010%5E%7B-9%7D%29%5Ctimes%20%28148.0%5C%20%5Chat%20i-110.0%5C%20%5Chat%20j%29%5C%5C%3D%281.539%5Ctimes%2010%5E%7B-6%7D%5C%20%5Chat%20i-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20%5Chat%20j%29%5C%20N.)
Thus, the y-component of the electric force on this charge is ![F_y = -1.144\times 10^{-6}\ N.](https://tex.z-dn.net/?f=F_y%20%3D%20-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20N.)
Both cells are formed in bone marrow.....but t cells matures into thymus....and b cells matures into bone marrow ! both helps in defense !
B cells can connect to antigens right on the surface of the invading virus or bacteria.
T- cells can only connect to virus antigens on the outside of infected cells.
for more info , comment !
Answer:
n = 1.76
Explanation:
According to the rule of ( n1 sin theta1 = n2 sin theta2 )
we know both angles so we insert them to the law and apply n1 = 1
so 1/2 = n2 sin 62 and we get the final answer
Answer:
The angular velocity of the propeller is 2.22 rad/s.
Explanation:
The angular velocity (ω) of the propeller is:
Where:
θ: is the angular displacement = 10.6 revolutions
t: is the time = 30 s
![\omega = \frac{\Delta \theta}{\Delta t} = \frac{10.6 rev*\frac{2\pi rad}{1 rev}}{30 s} = 2.22 rad/s](https://tex.z-dn.net/?f=%20%5Comega%20%3D%20%5Cfrac%7B%5CDelta%20%5Ctheta%7D%7B%5CDelta%20t%7D%20%3D%20%5Cfrac%7B10.6%20rev%2A%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%7D%7B30%20s%7D%20%3D%202.22%20rad%2Fs%20)
Therefore, the angular velocity of the propeller is 2.22 rad/s.
I hope it helps you!
Answer:
D. Top is emission; bottom absorption.
Explanation:
Emission and spectrum of elements are due to the element absorbing or emitting wavelength of e-m energy. Elementary particles of elements can absorb energy from a ground state to enter an excited state, creating an absorption spectrum, or they can lose energy and fall back to a lower energy state, creating an emission spectrum. A simple rule to differentiate between an emission and an absorption spectrum is that: "all absorbed wavelength is emitted, but not all emitted wavelength is absorbed."
From the image, the lines indicates wavelengths. We can see that all of the wavelengths of the bottom absorption spectrum coincides with some of the wavelength of the upper emission wavelengths.