Answer:
(A) = 2.13 s
(B) = 0.47 Hz
(C) = 0.25 m
(D) = 0.74 m/s
Explanation:
number of oscillations (n) = 15
time (t) = 32 secs
start point (L1) = 13 cm = 0.13 m
End point (L2) = 63 cm = 0.63 m
(A) period = time / number of oscillation
= 32 / 15 = 2.13 s
(B) frequency = 1 / period
= 1 / 2.13 = 0.47 Hz
(C) Amplitude = 0.5 ( L2 - L1 )
= 0.5 ( 0.63 - 0.13 )
= 0.25 m
(D) max speed = (2π / T) x A
= (2π / 2.13) x 0.25
= 0.74 m/s
Answer:
The remaining lights would shine with the same brightness.
Explanation:
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
Answer:
2500miles
Explanation:
Given parameters:
Time of travel = 5hrs
Average speed = 500miles/hr
Unknown:
Distance between the two cities = ?
Solution:
To solve this problem, we must understand that speed is the distance covered with time.
So;
Distance = speed x time
Distance = 500 x 5 = 2500miles
<span>positive acceleration involves speeding up, while negative acceleration involves slowing down</span>