Answer:
If a crest formed by one wave interferes with a trough formed by the other wave then the rope will not move at all.
Explanation:
Assume a straight rope tied to both ends is at rest. When a wave is created at one end of the rope, it travels to the other end of the rope through formation of alternative crest and trough. Due to these crest and trough the rope shifts up and down.
But when there are two waves travelling through the rope and both have opposite direction (directed towards one another) in such a way that crest formed by one wave is interfering with the trough formed by the other wave then due to this interference the waves will cancel the effects of each other on the rope and rope will be stable.
Answer:
0.9432 m/s
Explanation:
We are given;
Mass of swimmer;m_s = 64.38 kg
Mass of log; m_l = 237 kg
Velocity of swimmer; v_s = 3.472 m/s
Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.
So;
Initial momentum = final momentum
m_l × v_l = m_s × v_s
Where v_l is speed of the log relative to water
Making v_l the subject, we have;
v_l = (m_s × v_s)/m_l
Plugging in the relevant values, we have;
v_l = (64.38 × 3.472)/237
v_l = 0.9432 m/s
Answer:
The answer is explained below.
Explanation:
All the point on the disk has same angular acceleration. Here, the point P is at the midway between the center and the rim of the disk and the point Q is at rim of the disk.
So, the distance of the point Q from the axis is twicee the distance of the point P from the axis.
<em>Rp - R</em>
<em>Rq - 2R</em>
The linear acceleration is
α2 - Rα
So, the linear acceleration of Q is twice as great as the linear acceleration of P.
The speed of the particle when it is in the circular motion depends on the radius of the particle.
In this case, the speed of point Q is twice the speed of point P.
H=6cm(height of the object)
D=8cm(distance of the object from the lens)
d=4cm(distance of the image from the lens)
-----------
h=?(height of the image)
(M - magnification)
M = H/D = h/d
H×d=D×h
h=H×d/D
h=6×4/8
h=3cm
The height of the image is 3cm.
