Answer:
(a) x=ASin(ωt+Ф₀)=±(√3)A/2
(b) x=±(√2)A/2
Explanation:
For part (a)
V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)
Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3
x=ASin(ωt+Ф₀)=±(√3)A/2
For part(b)
U=0.5E and U+K=E→K=0.5E
E=K(Max)
(1/2)mv²=(0.5)(1/2)m(Vmax)²
V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4
x=±(√2)A/2
The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>
Why?
Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.
We can calculate the vertical force using the following formula:

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>
Have a nice day!
Answer:
flagpole
Explanation:
if it is about electricity then its flagpole
Therefore, it can be reasonably concluded according to your
unfinished syllogism, that there are many people who do not
think scientifically.
Answer:
f3 = 102 Hz
Explanation:
To find the frequency of the sound produced by the pipe you use the following formula:

n: number of the harmonic = 3
vs: speed of sound = 340 m/s
L: length of the pipe = 2.5 m
You replace the values of n, L and vs in order to calculate the frequency:

hence, the frequency of the third overtone is 102 Hz