Answer:
![N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons](https://tex.z-dn.net/?f=N_%7Belectrons%7D%3DQ_%7Btransfered%7D%2Fq_%7Belectron%7D%3D5.94%2A10%5E%7B18%7Delectrons)
Explanation:
The total charge is distributed over the two objects:
![Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\](https://tex.z-dn.net/?f=Q_%7Btotal%7D%2F2%3D%283.8%2A10%5E%7B-6%7DC%2B1.9C%29%2F2%3D0.9500019C%5C%5C)
The plate and the rod must have
. So the charge transferred from the plate to the rod is:
![Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\](https://tex.z-dn.net/?f=Q_%7Btransfered%7D%3D3.8%2A10%5E%7B-6%7DC-Q_%7Btotal%7D%2F2%3D3.8%2A10%5E%7B-6%7DC-0.9500019C%3D-0.9499981C%5C%5C)
Number of electrons:
![N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons](https://tex.z-dn.net/?f=N_%7Belectrons%7D%3DQ_%7Btransfered%7D%2Fq_%7Belectron%7D%3D-0.9499981C%2F%28-1.6%2A10%5E%7B-19%7DC%29%3D5.94%2A10%5E%7B18%7Delectrons)
Answer:
The recoil velocity is 0.819 m/s.
Explanation:
Given that,
Mass of ball m= 7.5 kg
Velocity v= 8.85 m/s
Mass of performer m'= 73.5 kg
Suppose if the performer is on nearly friction less roller skates, we need to calculate his recoil velocity
We need to calculate the recoil velocity
Using law of conservation momentum
![mv=(m+m')v'](https://tex.z-dn.net/?f=mv%3D%28m%2Bm%27%29v%27)
Where, m = mass of ball
m' = mass of performer
v = velocity of ball
Put the value into the formula
![v'=\dfrac{mv}{m+m'}](https://tex.z-dn.net/?f=v%27%3D%5Cdfrac%7Bmv%7D%7Bm%2Bm%27%7D)
Put the value into the formula
![v'=\dfrac{7.5\times8.85}{7.5+73.5}](https://tex.z-dn.net/?f=v%27%3D%5Cdfrac%7B7.5%5Ctimes8.85%7D%7B7.5%2B73.5%7D)
![v'=0.819\ m/s](https://tex.z-dn.net/?f=v%27%3D0.819%5C%20m%2Fs)
Hence, The recoil velocity is 0.819 m/s.
Answer:
The time taken for the sound wave to make the round trip is 0.16 s.
Explanation:
Given;
distance traveled by the sound wave, d = 120 m
bulk modulus of sea water, B = 2.3 x 10⁹ N/m²
density of sea water, ρ = 1022 kg
The speed of the wave is given by;
![v = \sqrt{\frac{B}{\rho} } \\\\v = \sqrt{\frac{2.3*10^9}{1022} }\\\\v = 1500.16 \ m/s](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BB%7D%7B%5Crho%7D%20%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2.3%2A10%5E9%7D%7B1022%7D%20%7D%5C%5C%5C%5Cv%20%3D%201500.16%20%5C%20m%2Fs)
Speed is given by;
![Speed = \frac{Distance}{Time}](https://tex.z-dn.net/?f=Speed%20%3D%20%5Cfrac%7BDistance%7D%7BTime%7D)
total distance of the round trip = 2 x 120m = 240 m
Time taken for the sound wave to make the round trip is given by;
![Time = \frac{Distance}{Speed} \\\\Time = \frac{240}{1500.16} \\\\Time = 0.16 \ second](https://tex.z-dn.net/?f=Time%20%3D%20%5Cfrac%7BDistance%7D%7BSpeed%7D%20%5C%5C%5C%5CTime%20%3D%20%5Cfrac%7B240%7D%7B1500.16%7D%20%5C%5C%5C%5CTime%20%3D%200.16%20%5C%20second)
Therefore, the time taken for the sound wave to make the round trip is 0.16 s.
For the answer to the question above,
the distance from i to j is 5 parts
(2 parts from i to k and 3 parts from k to j)
The y distance from i to j is
10 - 2 = 8
Each part is 8/5 = 1.6
Therefore the distance between the 2 parts from i to k is 3.2
From the y coordinate of I which is 2 plus the 3.2 to point k
2 + 3.2 = 5.2
Answer y =5.2
Now just convert that to fraction and that will be the answer