Answer:
9.7 ms⁻¹
Explanation:
The formula to find the acceleration is :
![Acceleration=\frac{Final \:Velocity-Initial\:Velocity}{Time}](https://tex.z-dn.net/?f=Acceleration%3D%5Cfrac%7BFinal%20%5C%3AVelocity-Initial%5C%3AVelocity%7D%7BTime%7D)
According to the question, we are asked to find the Final Velocity.
For that, let the Final Velocity be x.
![Acceleration=\frac{Final \:Velocity-Initial\:Velocity}{Time}](https://tex.z-dn.net/?f=Acceleration%3D%5Cfrac%7BFinal%20%5C%3AVelocity-Initial%5C%3AVelocity%7D%7BTime%7D)
![2m/s^{2} =\frac{x-3.7m/s}{3 s}\\](https://tex.z-dn.net/?f=2m%2Fs%5E%7B2%7D%20%3D%5Cfrac%7Bx-3.7m%2Fs%7D%7B3%20s%7D%5C%5C)
![2 =\frac{x-3.7}{3 }\\](https://tex.z-dn.net/?f=2%20%3D%5Cfrac%7Bx-3.7%7D%7B3%20%7D%5C%5C)
![2*3=x-3.7](https://tex.z-dn.net/?f=2%2A3%3Dx-3.7)
![6=x-3.7](https://tex.z-dn.net/?f=6%3Dx-3.7)
![6+3.7=x](https://tex.z-dn.net/?f=6%2B3.7%3Dx)
![9.7=x](https://tex.z-dn.net/?f=9.7%3Dx)
Therefore,
Final velocity = 9.7ms⁻¹
Hope this helps you :-)
Let me know if you have any other questions :-)
It's 0 m/s because you're both moving on the same machine at the same speed, so it appears that they're not moving
Answer:
0.8 N
Explanation:
From coulomb's law,
Formula:
F = kqq'/r²........................ Equation 1
Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.
Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.
Constant: k = 9×10⁹ Nm²/C²
Substitute these values into equation 1
F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²
F = 1800×10⁻³/2.25
F = 1.8/2.25
F = 0.8 N
For us to awnser the question you must give us the details
Answer:
In Young’s double-slit experiment using monochromatic light, the interference pattern consists of a central <u>MAXIMUM</u>.
Explanation:
In Young's double slit method of experiment we know that on the screen we have light of monochromatic source.
Due to this double slit system when light reaches to the screen there exist a finite path difference on the screen and which will create phase difference between two light reaches on the screen
It is given as
(Path difference)
so here if path difference is zero due to two slits at the center of the screen
Then in that case
![\Delta \phi = \frac{2\pi}{\lambda}(0) = 0](https://tex.z-dn.net/?f=%5CDelta%20%5Cphi%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%280%29%20%3D%200)
so we will have maximum intensity at the center of the screen when light will reach at the center of the screen.