Answer:
Speed, v = 312.34 m/s
Explanation:
The equation that describes a transverse wave on the string is given by :
![y=0.0120\ msin[(927\ rad/s)t-(3\ rad/m)x]](https://tex.z-dn.net/?f=y%3D0.0120%5C%20msin%5B%28927%5C%20rad%2Fs%29t-%283%5C%20rad%2Fm%29x%5D)
..............(1)
Where
y = displacement of a string particle
x = position of the particle on the string
The wave is travelling in the +x direction. We have to find the speed of the wave.
The general equation of traverse wave is given by :
................(2)
On comparing equation (1) and (2) we get,
k = 3 rad/m
Since, 
..............(3)
Also, 
Since, 
...............(4)
Speed of the wave is the product of frequency and wavelength i.e.
Using equation (3) and (4), the speed of the wave can be calculated as :
v = 312.34 m/s
Hence, the speed of the transverse wave is 312.34 m/s
F = ma
F 1500N
m = 100 kg
1500 = 100 * a
Solve for a.
Posiible units of speed and velocity aare miles per hour
Q: A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula:
h(t)=100+45t−4.9t2
(a)
What is its height after 3 seconds?
(b)What is its velocity after 3 seconds?
Answer:
(a) 190.9 m.
(b) 15.6 m/s upward
Explanation:
Given:
h(t) = 100 + 45t - 4.9t²
The height after 3 seconds,
t = 3 s
Substitute the value of t in to the equation above.
h(3) = 100+45(3)-4.9(3)²
h(3) = 100+135-44.1
h(3) = 190.9 m
Therefore the height after 3 seconds = 190.9 m.
(b) Velocity after 3 seconds
The velocity is obtained by differentiating h(t) with respect to time
v = dh(t)/dt
dh(t)/dt = 45-9.8t
v = 45 - 9.8t ......................................... Equation 1
t = 3 s.
Substitute the value of t into the equation above,
v = 45 - 9.8(3)
v = 45- 29.4
v = 15.6 m/s
Thus the velocity after 3 seconds = 15.6 m/s upward
