Explanation:
As per the problem,
![\Delta U = (V_{B} - V_{A})(-q) > 0](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%28V_%7BB%7D%20-%20V_%7BA%7D%29%28-q%29%20%3E%200)
When q > 0 then -q is a negative charge . Since, change in potential energy (
) increases.
or,
> 0
or, ![V_{A} > V_{B}](https://tex.z-dn.net/?f=V_%7BA%7D%20%3E%20V_%7BB%7D)
Therefore, both positive and negative charge will move from
to
and as
so both of them move through a negative potential difference.
Thus, we can conclude that the true statements are as follows.
- The positively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
- The negatively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
The trip time taken by the box is 1.5 sec.
Explanation:
Given:
distance(d) = 30 m
Velocity(v) = 20 m/s
By using formula:
Velocity = ![\frac{distance}{time}](https://tex.z-dn.net/?f=%5Cfrac%7Bdistance%7D%7Btime%7D)
or, 20 = ![\frac{30 }{time}](https://tex.z-dn.net/?f=%5Cfrac%7B30%20%7D%7Btime%7D)
or, time(t) = ![\frac{30}{20}](https://tex.z-dn.net/?f=%5Cfrac%7B30%7D%7B20%7D)
∴ time (t) = 1.5 sec
Hence the the block took 1.5 sec to slide 30 m.
Cellulat telephone operate with radio waves, a form of Electromagnetic waves
Answer:
t = 376.99 s
Explanation:
We must solve this problem with the equations and kinematics, let's start by looking for the speed of the vehicle,
v = d / t
v = 7/40
v = 0.175 m / s
Since the speed e remains constant, we must find the length of the circle is
L = 2π r
L = 2π 21
L = 131.95 m
In the problem it does not specify clearly, but in general the curves of the road correspond to half a circle, so the length of the road is
L ’= L / 2
L ’= 131.95 / 2 = 65.97 m
as the speed is constant
t = L ’/ v
t = 65.97 / 0.175
t = 376.99 s