Answer : The speed of waves along the wire is, 1.2 m/s
Explanation :
Formula used :

where,
= frequency = 
= wavelength =
c = speed of wave = ?
Now put all the given value in the above formula, we get the speed of waves along the wire.


Thus, the speed of waves along the wire is, 1.2 m/s
Answer:
The length of the rod for the condition on the question to be met is 
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is 
The mass of each small bodies is 
The moment of inertia of the three-body system with respect to the described axis is 
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

Where
is the moment of inertia of the rod about the describe axis which is mathematically represented as

And
the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
![I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}](https://tex.z-dn.net/?f=I_m%20%20%3D%20%20%20m%20%2A%20%5B%5Cfrac%7BL%7D%20%7B2%7D%20%5D%5E2%20%3D%20%20m%2A%20%20%5Cfrac%7BL%5E2%7D%7B4%7D)
Thus 
Hence

=> ![I = [\frac{M}{12} + \frac{m}{2}] L^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20%20%20%5B%5Cfrac%7BM%7D%7B12%7D%20%20%2B%20%5Cfrac%7Bm%7D%7B2%7D%5D%20L%5E2)
substituting vales we have
![0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2](https://tex.z-dn.net/?f=0.929%20%20%20%3D%20%20%20%20%5B%5Cfrac%7B3.41%7D%7B12%7D%20%20%2B%20%5Cfrac%7B0.249%7D%7B2%7D%5D%20L%5E2)


Answer:i belive the answer is E = 1/2 mv^2
hope this helps plz mark me as braliest
Answer:
0.8m per second
Explanation:
we measure speed as distance over time
distance=8m
time=10s
therefore speed = 8/10
=0.8m per second
Answer:
Explanation:
Left block is on surface with higher inclination so it will go down . If T be tension
For motion of block A ,
net force = mgsin60 - (T + mg cos 60 x μ ) , μ is coefficient of friction .
ma = mgsin60 - T - mg cos 60 x .1
10a = 277.13 - T - 16
= 261.13 - T
T = 261.13 - 10a
For motion of block B
T - mg sin30 - mgcos30 x μ = ma
T- 160 - 27.71 = 10 a
261.13 - 10a - 160 - 27.71 = 10a
73.42 = 20a
a = 3.67 ft / s²
common acceleration = 3.67 ft / s²