The statement which is true of a wave that’s propagating along the pavement and girders of a suspension bridge is A. The wave is mechanical, with particles vibrating in a direction that is parallel to that of the wave, forming compressions and rarefactions.
Answer:
9.9 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
![v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 5+0^2}\\\Rightarrow v=9.9\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2as%2Bu%5E2%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2%5Ctimes%209.81%5Ctimes%205%2B0%5E2%7D%5C%5C%5CRightarrow%20v%3D9.9%5C%20m%2Fs)
If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.
Hence, the speed of the diver just before striking the water is 9.9 m/s
Answer:
C. Quadruple the intensity
Explanation:
The intensity of the sound is proportional to square of amplitude of the sound.
I ∝ A²
![\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}](https://tex.z-dn.net/?f=%5Cfrac%7BI_1%7D%7BA_1%5E2%7D%20%3D%20%5Cfrac%7BI_2%7D%7BA_2%5E2%7D%5C%5C%5C%5CI_2%20%3D%20%5Cfrac%7BI_1A_2%5E2%7D%7BA_1%5E2%7D)
When the given sound is twice loud as the initial value, then the new amplitude is twice the former.
A₂ = 2A₁
![I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7BI_1A_2%5E2%7D%7BA_1%5E2%7D%20%5C%5C%5C%5CI_2%20%3D%20%5Cfrac%7BI_1%282A_1%29%5E2%7D%7BA_1%5E2%7D%20%5C%5C%5C%5CI_2%20%3D%20%5Cfrac%7B4I_1A_1%5E2%7D%7BA_1%5E2%7D%5C%5C%5C%5C%20I_2%20%3D%204I_1)
Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity
Answer:
a) 16 N
b) 2.13 m/s²
Explanation:
Draw a free body diagram of the tv stand. There are four forces:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force P pulling right.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
The net force in the x direction is:
∑F = P − Nμ
∑F = P − mgμ
∑F = 25 N − (7.5 kg) (10 m/s²) (0.12)
∑F = 16 N
Net force equals mass times acceleration:
∑F = ma
16 N = (7.5 kg) a
a = 2.13 m/s²