Answer:
The time taken is 
Explanation:
From the question we are told that
The length of steel the wire is 
The length of the copper wire is 
The diameter of the wire is 
The tension is 
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

Where
is the time taken to transverse the steel wire which is mathematically represented as
![t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%20l_1%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here
is the density of steel with a value 
So
![t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%2031%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B8920%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)

And
is the time taken to transverse the copper wire which is mathematically represented as
![t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%20l_2%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho_c%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here
is the density of steel with a value 
So
![t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%2017%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B7860%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)

So



Although you walked (14+9)= 23 meters of distance, you ended up only 16.64 meters from where you started. That's your "displacement".
Velocity = (displacement / (time)
Velocity = (16.64m)/(180s)
Velocity = 0.0925 m/s roughly Southwest.
(That's about 3.6 INCHES per second. Apparently, you really NEED that mocha java and it's associated drug.)
Answer:
A.
B.s=397.6 m
Explanation:
Given that
speed u= 284.4 m/s
time t = 1.4 s
here he want to reduce the velocity from 284.4 m/s to 0 m/s.
So the final speed v= 0 m/s
We know that
v= u + at
So now by putting the values
0 = 284.4 -a x 1.4 (here we take negative sign because this is the case of de acceleration)

So the acceleration while stopping will be
.
Lets take distance travel before come top rest is s
We know that


s=397.6 m
So the distance travel while stopping is 397.6 m.
Answer: <u>To establish a point from which to measure a future distance, or if movement has occurred.</u>
Explanation:
A reference point is a point with respect to which one comes to know whether any movement has occurred or not. A still point is required. When the object changes distance from that point, it means that object has moved. The distance can be calculated from one point to another.
Therefore, A reference point is used to establish a point from which future distance or if movement has occurred is detected.
Answer:
2 kg
Explanation:
Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.
Sum of the moments at the balance point:
-(Mg)(L/3) + (mg)(L/2 − L/3) = 0
(Mg)(L/3) = (mg)(L/2 − L/3)
(Mg)(L/3) = (mg)(L/6)
2M = m
M = 1 kg, so m = 2 kg.
The mass of the rod is 2 kg.