E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
(r + h)²
where,
k = 9 × 10^9Nm²C^-2
Q = total charge, 300uC = 300 × 10^ -6C
r = 8 × 10^ -2m
h = 16 × 10^ -2m
then,
E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>
(8e^-2 + 16e^-2)²
E = 4687500N/C
Earth's atmosphere blocks many types of light including gamma, x-rays most ultraviolet and infrared. So optical telescopes that use visible light and ultraviolet telescopes that are used to study very hot stars are much less effective on Earth.
Answer:
![\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B%28R%5E2%5Comega_o%7D%7B%28R%5E2%20%2B%20r%5E2%29%7D)
Explanation:
As we know that there is no external torque on the system of two disc
then the angular momentum of the system will remains conserved
So we will have
![L_i = L_f](https://tex.z-dn.net/?f=L_i%20%3D%20L_f)
now we have
![L_i = (\frac{1}{2}MR^2)\omega_o](https://tex.z-dn.net/?f=L_i%20%3D%20%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%29%5Comega_o)
also we have
![L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega](https://tex.z-dn.net/?f=L_f%20%3D%20%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DMr%5E2%29%5Comega)
now from above equation we have
![(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%29%5Comega_o%20%20%3D%20%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DMr%5E2%29%5Comega)
now we have
![\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7BMR%5E2%5Comega_o%7D%7B%28MR%5E2%20%2B%20Mr%5E2%29%7D)
![\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B%28R%5E2%5Comega_o%7D%7B%28R%5E2%20%2B%20r%5E2%29%7D)
This question is incomplete, the complete question is;
Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.
For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror
Answer: angle of incidence is 39.4°
Explanation:
Given that;
two plain mirrors intersect at right angle (90°)
distance d = 11.5 cm
S = 28.0 cm
Now the angle that the reflection ray males with first the mirror equal theta (∅)
so
tan∅ = (S/2) / d
tan∅ = (28/2) / 11.5
tan∅ = 14 / 11.5
tan∅ = 1.2173
∅ = tan⁻¹ (1.2173)
∅ = 50.6°
so angle of incidence = 90° - ∅
= 90° - 50.6°
= 39.4°
Therefore angle of incidence is 39.4°
Answer:
13.4 m/s^2
Explanation:
40 divided by 3 equals 13.3 repeating.