Answer:
The air-water interface is an example of<em> </em>boundary. The <u><em>transmitted</em></u><em> </em> portion of the initial wave energy is way smaller than the <u><em>reflected</em></u><em> </em> portion. This makes the <u><em>boundary</em></u> wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can <u><em>travel directly to your ear</em></u>.
Explanation:
The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.
Answer:
Explanation:
The minimum magnitude of acceleration = 3 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 3 x 1²
= - 3 + 1.5
= - 1.5 m
position at t = 1 s
= 10 - 1.5
= 8.5 m
The maximum magnitude of acceleration = 6 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 6 x 1²
= - 3 + 3
= 0
position at t = 1 s
= 10 +0
= 10 m
So range of position is 8.5 m to 10 m .
Answer:
Option B and C are True
Note: The attachment below shows the force diagram
Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂;
w = T₁/cos 60° -----(1)
w = T₂/cos 30° ----(2)
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°
T₁ cos 30° = T₂ cos 60°
T₂/T₁ = cos 30°/cos 60°
T₂/T₁ =1.73
Therefore, option a is false since T₂ > T₁
Option B is true since T₁ cos 30° = T₂ cos 60°
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
The problem is about magnitude of the displacement vector of the lady bug and its directions. so the magnitude of the displacement of the lady bug is 16 in / 2 because it started from the center so the magnitude is 8 in. and the direction is the rotation of the turn table which is 60 degrees
Answer:T=116.84 N
Explanation:
Given
Weight of hiker =1040 N
acceleration 
Force exerted by Rope is equal to Tension in the rope
