How much is a ball that is tethered to a post accelerating if its linear speed is 3.5 m/s and the ball is 0.91 meters away from
1 answer:
Answer:
Explanation:
The motion of the tethered ball around the pole can be modeled as uniform circular motion. The acceleration of the objects executing uniform circular motion is due to the change in direction of their velocity. This is called centripetal acceleration. It is given by the following formula:
where,
ac = centripetal acceleration = ?
v = speed of ball = 3.5 m/s
r = distance of ball from center of rotation = 0.91 m
Using these values in equation, we get:
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Answer:
q = 13 C
Explanation:
Given that,
Electric current, I = 20 A
The starter motor runs for 0.65 s before the car engine starts up.
We need to find how much of electric charge passes through the starter motor. Let it is q.
Electric current = electric charge/time
So, 13 C of electric charge passes through the starter motor.
Answer:
0.015
Explanation:
Total volume of water coming out = 1.33
Also volume = Cross sectional area*Length covered
Length covered = Velocity *time
=24.5*3.55
=86.97 meter
Let the cross sectional area be A.
1.33 = 86.97*A
A =0.015
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V