A copper wire has a square cross section 2.1 mm on a side. The wire is 4.3 m long and carries a current of 3.6 A. The density of
free electrons is 8.5 * 1028>m3 . Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?
1 answer:
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The Newton's second law and kinematics allows to find the result of the error when placing the masses in the car is:
- Acceleration decreases.
- The plot of position vs. time squared is still linear, but the slope is less.
Newton's second law gives a relationship between the net force, mass and acceleration of the body
F = ma
a =
Where F is the force, m the mass and the acceleration
Kinematics studies the motion of bodies, looking for relationships between position, velocity, and acceleration.
x = v₀ t + ½ a t²
In the experiment, the position and time of the car are measured, starting from rest in each experiment.
x = ½ a t²
Let's substitute
x = ½ ( ) t²
This is the equation that the students should graph, a graph of the position versus time squared can be made to obtain a line.
The slope of the line is the acceleration of the car, which is related to the force that is the weight of the hanging mass.
When the student makes the mistake of placing the mass on the car, the acceleration decreases therefore in a graph the position values are smaller for each time.
The linearity of the graph is maintained, but when calculating the slope it gives lower values.
In conclusion using Newton's second law and kinematics we can find the result of the error when placing the masses on the car is:
- Acceleration decreases.
- The plot of position vs. time squared is still linear, but the slope is less.
Learn more here: brainly.com/question/11298125
Answer:
0(zero)
Explanation:
Given that =0.20kg, radius of the bowl,
and that the mass slides from point A to B.
-Since the normal force is perpendicular to the displacement at every point of the motion,then work done by the normal force is zero.