Answer:
the tension of the rope is 34.95 N
Explanation:
Given;
length of the rope, L = 3 m
mass of the rope, m = 0.105 kg
frequency of the wave, f = 40 Hz
wavelength of the wave, λ = 0.79 m
Let the tension of the rope = T
The speed of the wave is given as;
Therefore, the tension of the rope is 34.95 N
I believe that the correct answer you are looking for is the distance traveled
<span>So we want to know what statement is an accurate description of vibrations. So humans can hear sound frequencies from 20-20000 Hz. Below 20 Hz is infra sound and above 20000 Hz is ultra sound. Humans cant hear both infra sound and ultra sound so the correct answer is A.</span>
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
2.c
3.b
1.a
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