Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 × 
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 × )²×ω²× 34.64 ) / 2
0.050 ×(6.43 × )²×ω²× 34.64 = 108
ω² = 108 / 7.160 × 
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
The answer should be C) Slide against each other.
I believe it would be 2m/s.
Answer:
a = 2 [m/s²]
Explanation:
To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.
We can use the following equation.
where:
Vf = final velocity = 11 [m/s]
Vo = initial velocity = 5 [m/s]
a = acceleration [m/s²]
t = time = 3 [s]
![11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]](https://tex.z-dn.net/?f=11%20%3D%205%20%2B%20a%2A3%5C%5C6%3D3%2Aa%5C%5Ca%3D%202%5Bm%2Fs%5E%7B2%7D%20%5D)
Answer:
The final velocity of the vehicle is 10.39 m/s.
Explanation:
Given;
acceleration of the vehicle, a = 2.7 m/s²
distance moved by the vehicle, d = 20 m
The final velocity of the vehicle is calculated using the following kinematic equation;
v² = u² + 2ah
v² = 0 + 2 x 2.7 x 20
v² = 108
v = √108
v = 10.39 m/s
Therefore, the final velocity of the vehicle is 10.39 m/s.
