Answer:
The frequency of these waves is ![4.27\times10^{-2}\ Hz](https://tex.z-dn.net/?f=4.27%5Ctimes10%5E%7B-2%7D%5C%20Hz)
Explanation:
Given that,
Wavelength = 6.6 km
Distance = 8810 km
Time t = 8.67 hr
We need to calculate the velocity of sound
Using formula of velocity
![v = \dfrac{D}{T}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7BD%7D%7BT%7D)
Where, D = distance
T = time
Put the value into the formula
![v =\dfrac{8810}{8.67}](https://tex.z-dn.net/?f=v%20%3D%5Cdfrac%7B8810%7D%7B8.67%7D)
![v=1016\ km/hr](https://tex.z-dn.net/?f=v%3D1016%5C%20km%2Fhr)
We need to calculate the frequency
Using formula of frequency
![v=n\lambda](https://tex.z-dn.net/?f=v%3Dn%5Clambda)
![n=\dfrac{v}{\lambda}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7Bv%7D%7B%5Clambda%7D)
Put the value into the formula
![n=\dfrac{1016}{6.6}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B1016%7D%7B6.6%7D)
![n=153.93\ hr](https://tex.z-dn.net/?f=n%3D153.93%5C%20hr)
![n=\dfrac{153.93}{60\times60}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B153.93%7D%7B60%5Ctimes60%7D)
![n=0.0427\ Hz](https://tex.z-dn.net/?f=n%3D0.0427%5C%20Hz)
![n=4.27\times10^{-2}\ Hz](https://tex.z-dn.net/?f=n%3D4.27%5Ctimes10%5E%7B-2%7D%5C%20Hz)
Hence, The frequency of these waves is ![4.27\times10^{-2}\ Hz](https://tex.z-dn.net/?f=4.27%5Ctimes10%5E%7B-2%7D%5C%20Hz)
The answer would be 187.95 kg.m/s.
To get the momentum, all you have to do is multiply the mass of the moving object by the velocity.
p = mv
Where:
P = momentum
m = mass
v = velocity
Not the question is asking what is the total momentum of the football player and uniform. So we need to first get the combined mass of the football player and the uniform.
Mass of football player = 85.0 kg
Mass of the uniform = <u> 4.5 kg</u>
TOTAL MASS 89.5 kg
So now we have the mass. So let us get the momentum of the combined masses.
p = mv
= (89.5kg)(2.1m/s)
= 187.95 kg.m/s
In order for an object to move, the forces acting on it must be ''unbalanced forces'' because balanced forces means equal canceling each other out so 0 to 0 while unbalanced means one side or thing is higher or lower than another and that makes objects move which are unbalanced since a side is not equal to the other.
The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:
![p_A + p_B = p'_A + p'_B](https://tex.z-dn.net/?f=p_A%20%2B%20p_B%20%3D%20p%27_A%20%2B%20p%27_B)
where:
is the momentum of block A before the collision
is the momentum of block B before the collision
is the momentum of block A after the collision
is the momentum of block B after the collision
Solving for
, we find:
![p'_B = p_A + p_B - p'_A = -100 +(-150) - (-200)=-50 kg m/s](https://tex.z-dn.net/?f=p%27_B%20%3D%20p_A%20%2B%20p_B%20-%20p%27_A%20%3D%20-100%20%2B%28-150%29%20-%20%20%28-200%29%3D-50%20kg%20m%2Fs)
So, the momentum of block B after the collision is -50 kg m/s.
Learn more about momentum:
brainly.com/question/7973509
brainly.com/question/6573742
brainly.com/question/2370982
brainly.com/question/9484203
#LearnwithBrainly