Answer:
The density of a material affects the speed that a wave will be transmitted through it. In general, the denser the transparent material, the more slowly light travels through it.
Answer: 109.4 mm
Explanation: <u>Distance</u> is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time: d = vt
The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:
location of collision =
= 0.0525 m
To reach that point, both steamrollers will have spent



t = 0.04375 s
The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:
d = 2.5*0.04375
d = 0.109375 m
Until it is crushed, the fly will have traveled 109.4 mm.
Answer:
.
Explanation:
Since no external force is acting on the system.
Therefore, Total energy remains constant before and after.
So, Total energy of system= energy due to potential applied+kinetic energy

(Here v=velocity ,V=potential ,q=charge and m=mass).
Putting values .
We get,
.
At point B charged particle is moving faster as compared to point A.
Hence, it is the required solution.
Answer:
a) 1.6*10^6 V
b) 13.35*10^6 V
Explanation:
The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:
(1)
q1 = 3.90µC = 3.90*10^-6 C
q2 = -2.4µC = -2.4*10^-6 C
r1 = 1.25 cm = 0.0125 m
r2 = -1.80 cm = -0.018 m
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
You replace all the parameters in the equation (1):
![V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V](https://tex.z-dn.net/?f=V%3Dk%5B%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%5D%5C%5C%5C%5CV%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0125m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.018m%7D%5D%3D1.6%2A10%5E6V)
hence, the total electric potential is approximately 1.6*10^6 V
b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:
r1 = 0.0150m - 0.0125m = 0.0025m
r2= 0.015m + 0.018m = 0.033m
Then, you replace in the equation (1):
![V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V](https://tex.z-dn.net/?f=V%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0025m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.033m%7D%5D%3D13.35%2A10%5E6V)
hence, for y = 1.50cm you obtain V = 13.35*10^6 V
Answer:
Stellar Spectra Classification The patterns of lines detected in stellar spectra are used by astronomers to classify stars into spectral classes. These spectral classes are a measure of a star's surface temperature since the temperature of a star dictates which absorption lines are present in its spectrum.
Explanation: