Answer:
distance between both oasis ( 1 and 2) is 27.83 Km
Explanation:
let d is the distance between oasis1 and oasis 2
from figure
OC = 25cos 30
OE = 25sin30
OE = CD
Therefore BC = 30-25sin30
distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem
in![\Delta BCO](https://tex.z-dn.net/?f=%5CDelta%20BCO)
![OB^2 = BC^2 + OC^2](https://tex.z-dn.net/?f=OB%5E2%20%3D%20BC%5E2%20%2B%20OC%5E2)
PUTTING ALL VALUE IN ABOVE EQUATION
![d^2 = 930-25sin30)^2 + (25cos30)^2](https://tex.z-dn.net/?f=d%5E2%20%3D%20930-25sin30%29%5E2%20%2B%20%2825cos30%29%5E2)
![d^2 = 775](https://tex.z-dn.net/?f=d%5E2%20%3D%20775)
d = 27.83 Km
distance between both oasis ( 1 and 2) is 27.83 Km
Initially, the velocity vector is
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by
, so the velocity is
.
Converting back to direction and magnitude, we get ![\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle](https://tex.z-dn.net/?f=%5Clangle%20r%2C%5Ctheta%20%5Crangle%3D%5Clangle%20%5Csqrt%7B29.11%5E2%2B24.23%5E2%7D%2Ctan%5E%7B-1%7D%28%5Cfrac%7B29.11%7D%7B24.23%7D%29%20%5Crangle%20%3D%20%5Clangle%2037.87%2C50.2%5E%7B%5Ccirc%7D%5Crangle)
Answer:
![F_2 = 29.54 N](https://tex.z-dn.net/?f=F_2%20%3D%2029.54%20N)
Explanation:
As we know that the combination is maintained at rest position
So we will take net torque on the system to be ZERO
so we know that
![\tau = \vec r \times \vec F](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cvec%20r%20%5Ctimes%20%5Cvec%20F)
here we will have
![\vec r_1 \times F_1 = \vec r_2 \times F_2](https://tex.z-dn.net/?f=%5Cvec%20r_1%20%5Ctimes%20F_1%20%3D%20%5Cvec%20r_2%20%5Ctimes%20F_2)
so we have
![13 \times 50 = 22 \times F_2](https://tex.z-dn.net/?f=13%20%5Ctimes%2050%20%3D%2022%20%5Ctimes%20F_2)
so we have
![F_2 = \frac{13 \times 50}{22}](https://tex.z-dn.net/?f=F_2%20%3D%20%5Cfrac%7B13%20%5Ctimes%2050%7D%7B22%7D)
![F_2 = 29.54 N](https://tex.z-dn.net/?f=F_2%20%3D%2029.54%20N)
Answer:
The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.
F = N e E where E is the value of the field and N e the charge Q
M g = N e E and M g is the weight of the drop
N = M g / (e E)
N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13
.00011 kg is a very large drop
Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs
Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons