Echo sounding is a type of SONAR used to determine the depth of water by transmitting sound pulses into water. The time interval between emission and return of a pulse is recorded, which is used to determine the depth of water along with the speed of sound in water at the time.
Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force ![F=14.413\ N](https://tex.z-dn.net/?f=F%3D14.413%5C%20N)
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem
![r=\sqrt{x_{2}^2+y_{2}^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7Bx_%7B2%7D%5E2%2By_%7B2%7D%5E2%7D)
Put the value into the formula
![r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%288.2%5Ctimes10%5E%7B-2%7D%29%5E2%2B%283.1%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![r=0.0876\ m](https://tex.z-dn.net/?f=r%3D0.0876%5C%20m)
We need to calculate the magnitude of the charge q₃
Using formula of net force
![F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})](https://tex.z-dn.net/?f=F_%7B12%7D%3Dkq_%7B2%7D%28%5Cdfrac%7Bq_%7B3%7D%7D%7Br_%7B3%7D%5E2%7D%2B%5Cdfrac%7Bq_%7B1%7D%7D%7Br_%7B1%7D%5E2%7D%29)
Put the value into the formula
![14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})](https://tex.z-dn.net/?f=14.413%3D9%5Ctimes10%5E%7B9%7D%5Ctimes7.7%5Ctimes10%5E%7B-6%7D%28%5Cdfrac%7Bq_%7B3%7D%7D%7B%280.0438%29%5E2%7D%2B%5Cdfrac%7B-3%5Ctimes10%5E%7B-6%7D%7D%7B%280.0876%29%5E2%7D%29)
![(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}](https://tex.z-dn.net/?f=%28%5Cdfrac%7Bq_%7B3%7D%7D%7B%284.38%5Ctimes10%5E%7B-2%7D%29%5E2%7D%2B%5Cdfrac%7B-3%5Ctimes10%5E%7B-6%7D%7D%7B%280.0876%29%5E2%7D%29%3D%5Cdfrac%7B14.413%7D%7B9%5Ctimes10%5E%7B9%7D%5Ctimes7.7%5Ctimes10%5E%7B-6%7D%7D)
![\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}](https://tex.z-dn.net/?f=%5Cdfrac%7Bq_%7B3%7D%7D%7B%280.0438%29%5E2%7D%3D207%5Ctimes10%5E%7B-4%7D%2B3.909%5Ctimes10%5E%7B-4%7D)
![q_{3}=0.0210909\times(0.0438)^2](https://tex.z-dn.net/?f=q_%7B3%7D%3D0.0210909%5Ctimes%280.0438%29%5E2)
![q_{3}=40.46\times10^{-6}\ C](https://tex.z-dn.net/?f=q_%7B3%7D%3D40.46%5Ctimes10%5E%7B-6%7D%5C%20C)
![q_{3}=40.46\ \mu C](https://tex.z-dn.net/?f=q_%7B3%7D%3D40.46%5C%20%5Cmu%20C)
Hence, The value of charge q₃ is 40.46 μC.
To contrast inner and outer planets we will start with the climate of the planets and then move on to there lighting. To start the planets closet to the sun, mercury, venus, earth and mars, are all hot compared to the further one, jupiter, saturn, uranus, neptune. This distance also makes the farthe away planets darker than the ones closer. Now to compare all the planets vary from either gass or solid, rocky or icy. All of them spin around the sun and all have objects spinning around them, moons.
Answer:
d. 50 C
Explanation:
In this problem, we have to add 800 ml of water at 20 Celsius to 800 ml of water at 80 Celsius.
According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.
The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:
Final temperature = (20 C + 80 C)/2
= 50 Celsius
Answer:
Immunization, or immunisation, is the process by which an individual's immune system becomes fortified against an infectious agent.
Explanation:
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPS</em><em> </em>
<em>HAVE</em><em> </em><em>A</em><em> </em><em>NICE</em><em> </em><em>DAY</em><em> </em><em>:)</em><em> </em>
<em>XXITZFLIRTYQUEENXX</em><em> </em>