The light coming out of a concave lens will never meet.
So, the answer is A. will never meet.
Happy Studying! ^^
Explanation:
change 0.5 g to kg so 0.005kg then change 100 ml to m so 0.001m so density=mass over volume so from there you can continue
Answer:
ω = 0.05 rad/s
Explanation:
We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,
![Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\](https://tex.z-dn.net/?f=Centripetal%20Force%20%3D%20Weight%5C%5C%5Cfrac%7Bmv%5E%7B2%7D%7D%7Br%7D%20%3D%20mg%5C%5C%5C%5Chere%2C%5C%5Cv%20%3D%20linear%5C%20speed%20%3D%20r%5Comega%20%5C%5Ctherefore%2C%5C%5C%5Cfrac%7B%28r%5Comega%29%5E%7B2%7D%7D%7Br%7D%20%3D%20g%5C%5C%5C%5C%5Comega%5E%7B2%7D%20%3D%20%5Cfrac%7Bg%7D%7Br%7D%5C%5C%5C%5C%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bg%7D%7Br%7D%7D%5C%5C)
where,
ω = angular velocity of cylinder = ?
g = required acceleration = 9.8 m/s²
r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m
Therefore,
![\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B9.8%5C%20m%2Fs%5E%7B2%7D%7D%7B4023.36%5C%20m%7D%7D%5C%5C%5C%5C)
<u>ω = 0.05 rad/s</u>
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5