Explanation:
As per the problem,
When q > 0 then -q is a negative charge . Since, change in potential energy (
) increases.
or, 
> 0
or, 
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).
Answer:
Explanation:
If the initial velocity is U
Then the horizontal component of the velocity is
Ux= Ucosθ
Then the range for a projectile is give as
R=Ux.t
Where t is the time of flight
The time of flight is given as
t=2USinθ/g
Therefore,
R=Ux.t
R=UCosθ.2USinθ/g
R=U^2×2SinθCosθ/g
Then, from trigonometric ratio
2SinθCosθ= Sin2θ
R=U^2Sin2θ/g
Given that θ=32° and g=9.81m/s^2
Then
R=U^2Sin2×32/9.81
R=U^2Sin64/9.81
R=0.0916U^2
Then, range is given by R=0.0916U^2
A=0.0916U^2.
T
The box is at a distance A from the point of projection. Then the range R=A
R=0.0916U^2
A=0.0916U^2
Then,
U^2=A/0.0916
U^2=10.915A
Then the initial velocity should be
U=√10.915A
U=3.3√A
(C)
Explanation:
The circle has a radius r = 0.5 m, which means that its circumference C is
One revolution means that the stopper travels a distance equal to the circumference of the circle so the velocity of the stopper is
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
F~1/r²
doubling the distance r, Decreases the force by ¼
