Explanation:
Given that,
The distance between two conducting plates is 7 cm.
The difference in potential between the plates is 560 V.
To find,
We need to tell is the positive or the negative plate at the higher potential?
Solution,
We know that the electric potential is defined as the work done in moving a test charge from one position to another. So, the positive plate is always at the higher potential.
Hence, this is the required solution.
In vacuum and empty space, it's 299,792,458 meters per second
(186,282.4 miles per second).
In any material stuff, it's somewhat less. How much less depends on
what stuff it is ... it's different in each material "medium".
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
The speed of light in that medium is
.
<u>Explanation:</u>
It is known that the light's speed is constant when it travels in vacuum and the value is
. When the light enters another medium other than vacuum, its speed get decreased as the light gets refracted by an angle.
The amount of refraction can be determined by the index of refraction or refractive index of the medium. The refraction index is measured as the ratios of speed of light in vacuum to that in the medium. It is represented as η = ![\frac {c}{v}](https://tex.z-dn.net/?f=%5Cfrac%20%7Bc%7D%7Bv%7D)
So, here η is the index of refraction of a medium which is given as 1.4, c is the light's speed in vacuum (
) and v is the light's speed in that medium which we need to find.
![1.4= \frac{(3 \times 10 ^ 8)} {v}](https://tex.z-dn.net/?f=1.4%3D%20%20%5Cfrac%7B%283%20%5Ctimes%2010%20%5E%208%29%7D%20%7Bv%7D)
![v= \frac {(3 \times 10^8)}{1.4} =2.14 \times 10^8 \ m/s](https://tex.z-dn.net/?f=v%3D%20%20%5Cfrac%20%7B%283%20%5Ctimes%2010%5E8%29%7D%7B1.4%7D%20%3D2.14%20%5Ctimes%2010%5E8%20%5C%20m%2Fs)
Thus the speed of light in that medium is ![2.14 \times 10^8 \ m/s.](https://tex.z-dn.net/?f=2.14%20%5Ctimes%2010%5E8%20%5C%20m%2Fs.)
Answer:
![T = 5.45 h](https://tex.z-dn.net/?f=T%20%3D%205.45%20h)
Explanation:
As we know that time period of one revolution around Mars is given as
![T = 2\pi\sqrt{\frac{r^3}{GM}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7Br%5E3%7D%7BGM%7D%7D)
here we know that
![r = 2.2 R](https://tex.z-dn.net/?f=r%20%3D%202.2%20R)
here we know that radius of Mars is given as
![R = 3.396 \times 10^6 m](https://tex.z-dn.net/?f=R%20%3D%203.396%20%5Ctimes%2010%5E6%20m)
mass of Mars is given as
![M = 6.42 \times 10^{23} kg](https://tex.z-dn.net/?f=M%20%3D%206.42%20%5Ctimes%2010%5E%7B23%7D%20kg)
now we have
![T = 2\pi \sqrt{\frac{(2.2\times 3.396 \times 10^6)^3}{(6.67 \times 10^{-11})(6.42\times 10^{23})}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7B%282.2%5Ctimes%203.396%20%5Ctimes%2010%5E6%29%5E3%7D%7B%286.67%20%5Ctimes%2010%5E%7B-11%7D%29%286.42%5Ctimes%2010%5E%7B23%7D%29%7D%7D)
![T = 19608 s](https://tex.z-dn.net/?f=T%20%3D%2019608%20s)
![T = 5.45 h](https://tex.z-dn.net/?f=T%20%3D%205.45%20h)