Answer:
The current through the tube is 73.39A.
Explanation:
The relationship between the resistivity , the electric field , and the current density is given by
This equation can be solved for to get:
Since the current is
Now, for the tube of mercury , , and the area is ; therefore,
Hence, the current through the mercury tube is 73.39A.
Answer:
microsecond = 1 × 10-6 seconds
1 second = 1 × 100 seconds
1 microsecond = (1 / 1) × 10-6 × 10-0 seconds
1 microsecond = (1) × 10-6-0 seconds
1 microsecond = (1) × 10-6 seconds
1 microsecond = 1 × 1.0E-6 seconds
1 microsecond = 1.0E-6 seconds,just do like it
Answer:
The answer is related to physics
Explanation:
In physics, rest is the state of an object being stationary relative to a particular frame of reference or another object; when the position of a body with respect to its surroundings does not change with time it is said to be "at rest". According to the theory of relativity, it is said that an object is "at rest relative to" another. For example, a train decelerates approaching a station and eventually comes to rest alongside the platform. The train can be said to be "at rest with respect to the station", or, as the correct frame of reference is usually implicit and/or provided by context, simply "at rest". In reality, there is nothing at absolute rest. For example, Earth's gravitation constantly pulls objects toward its surface, while Earth is one of the objects the Sun constantly pulls towards itself, causing it to orbit the Sun; the Sun, in turn, orbits the center of the Milky Way; and so on.
Answer:
Linear momentum is a property of objects which are changing their position with respect to a reference point.
Angular momentum is a property of objects which are changing the angle of their position vector with respect to a reference point.
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,