Answer:
True
Explanation:
If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell
According to Gauss law
∅ = EA =-Q/∈₀
Where ∅ is the electric flux through the gaussian surface and E is the electric field strength
If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero
Answer:
337k
Explanation:
First, let us find the difference between the given two temperatures.
Difference = 85°C - 21°C
= 64°C
<u>And now we have to write the temperature in kelvins.</u>
To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.
<u>Let us find it now.</u>
64°C + 273 = 337k
Therefore,
64°C ⇒ <u>337k</u>
At the ground the ball will always have velocity along the direction of gravity. If upward motion is taken positive it will always have negative velocity at the ground because, if the ball was given an initial upward velocity then gravity will decelerate it and bring it down with a negative final velocity. If the ball is given an initial downward velocity then the ball will be further accelerated by gravity in the downward direction only, again maintaining negative direction. The magnitude however in both cases will be different. the final velocity at the ground will have higher magnitude in case of elevator moving downwards.
(1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)
= (1.9 x 365.24 x 86,400 x 10⁹) nanosec
= 6.00 x 10¹⁶ nanoseconds
Answer: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below
