Answer:
The right sphere is negatively charged, the left sphere is charged positively.
Explanation:
When a negatively charged rod is held above the top of left sphere, the rod will attract positive charges and repel negative charges. As the sphere are initially touching each other so positive charges from the both spheres will moves toward the rod. When we separate the spheres positive charges from right sphere have already moved toward the rod i.e. left sphere, creating a deficiency of positive charges in the right sphere and excessiveness of positive charges in left sphere , hence the right sphere will remain negatively charged and left sphere will remain positively charged.
Answer:
<h2>4 m/s</h2>
Explanation:
The speed of the ball can be found by using the formula
![s= \frac{p}{m} \\](https://tex.z-dn.net/?f=s%3D%20%20%5Cfrac%7Bp%7D%7Bm%7D%20%20%5C%5C%20)
p is the momentum
m is the mass
From the question we have
![v = \frac{40}{10} = 4 \\](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B40%7D%7B10%7D%20%20%3D%204%20%5C%5C%20)
We have the final answer as
<h3>4 m/s</h3>
Hope this helps you
Answer:
n = 3
Solution:
Since, the slit used is same and hence slit distance 'x' will also be same.
Also, the wavelengths coincide,
will also be same.
Using Bragg's eqn for both the wavelengths:
![xsin\theta = n\lambda](https://tex.z-dn.net/?f=xsin%5Ctheta%20%3D%20n%5Clambda)
(1)
![xsin\theta = (n + 1)\lambda](https://tex.z-dn.net/?f=xsin%5Ctheta%20%3D%20%28n%20%2B%201%29%5Clambda)
(2)
equate eqn (1) and (2):
![n\times 680.0\times 10^{- 9} = (n + 1)\times 510.0\times 10^{- 9}](https://tex.z-dn.net/?f=%20n%5Ctimes%20680.0%5Ctimes%2010%5E%7B-%209%7D%20%3D%20%28n%20%2B%201%29%5Ctimes%20510.0%5Ctimes%2010%5E%7B-%209%7D)
![n = \frac{510.0\times 10^{- 9}}{680.0\times 10^{- 9} - 510.0\times 10^{- 9}}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B510.0%5Ctimes%2010%5E%7B-%209%7D%7D%7B680.0%5Ctimes%2010%5E%7B-%209%7D%20-%20510.0%5Ctimes%2010%5E%7B-%209%7D%7D)
n = 3
Answer:
The length of the wire is 2.86 m
Explanation:
Given:
Current
A
Magnetic field
T
Force
N
Angle between wire and magnetic field
46°
The magnetic force on the current carrying wire is given by,
![F = BIl\sin \theta](https://tex.z-dn.net/?f=F%20%3D%20BIl%5Csin%20%5Ctheta)
Where
length of wire
![l = \frac{F}{BI \sin 46}](https://tex.z-dn.net/?f=l%20%3D%20%5Cfrac%7BF%7D%7BBI%20%5Csin%2046%7D)
![l = \frac{9.4 \times 10^{-5} }{6 \times 10^{-5}\times 0.76 \times 0.7193}](https://tex.z-dn.net/?f=l%20%3D%20%5Cfrac%7B9.4%20%5Ctimes%2010%5E%7B-5%7D%20%7D%7B6%20%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20%200.76%20%5Ctimes%200.7193%7D)
m
Therefore, the length of the wire is 2.86 m
Answer:
0.050V
Explanation:
To solve this problem it is necessary to use the concepts related to the potential between two objects that have a magnetic field, this concept is represented in the equation.
![\int dV = \int_{0}^{l/2} Bv (dl)](https://tex.z-dn.net/?f=%5Cint%20dV%20%3D%20%5Cint_%7B0%7D%5E%7Bl%2F2%7D%20Bv%20%28dl%29)
Where,
v= tangencial velocity
B = Magnetic Field
We know for definition that,
![v= l\omega](https://tex.z-dn.net/?f=v%3D%20l%5Comega)
Where,
L = length
Angular velocity
We can replace this values in our first equation then,
![\int dV = \int_{0}^{l/2} B (l\omega) (dl)](https://tex.z-dn.net/?f=%5Cint%20dV%20%3D%20%5Cint_%7B0%7D%5E%7Bl%2F2%7D%20B%20%28l%5Comega%29%20%28dl%29)
Integrating we have,
![V = \frac{1}{8} Bl^2 \omega](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1%7D%7B8%7D%20Bl%5E2%20%5Comega)
Replacing the values,
![V= \frac{1}{8} (8*10^{-3})(12.56)(4)](https://tex.z-dn.net/?f=V%3D%20%5Cfrac%7B1%7D%7B8%7D%20%288%2A10%5E%7B-3%7D%29%2812.56%29%284%29)
![V = 0.050V](https://tex.z-dn.net/?f=V%20%3D%200.050V)
Therefore the potential difference between the center of the rod and the other rod is 0.050V