Answer:
I = 1.05x10⁻³ A
Explanation:
By definition, an electric current is the rate of charge flow at a given time:
Where:
q: is the electrons charge = 1.602x10⁻¹⁹ C
t: is the time
In a circular motion, the time is given by:
Where:
ω: is the angular speed = v/r
v: is the speed = 2.19x10⁶ m/s
r: is the radius = 5.29x10⁻¹¹ m
Now, the effective current is:
Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.
I hope it helps you!
I also agree to A. The force will always pull back a lot deals with gravitational pull
Answer:
1. E
2.XD
Explanation:
Matagal nako mongoloyd:)))
Let D be the total distance (say in meters) traveled by the train and T the time (say in seconds) it takes to do so. (Assume the train moves in a straight line in only one direction.) Then the average velocity of the train as it covers this distance is
v (ave) = D/T
We're told the train can traverse a distance of D/4 in a matter of T/2 seconds if it moves at a speed of 5 m/s. This means
D/4 = (5 m/s) (T/2)
⇒ 5 m/s = 1/2 D/T
⇒ v (ave) = D/T = 10 m/s
An object moving in a circle wants to move in a straight line.
The motion of an object is tangent to the circle.
Tangential speed remains constant for an object with uniform circular motion.
Only the direction of the object changes, which means velocity is changing and the object is accelerating