Potential energy=mgh
m=3kg
g=<span> 9.81 m/s)</span>
h=<span> 1.00 m
p.E=</span><span> 3*9.81*1 = 29.43 Joules
</span><span> (1/2)mv^2=mgh
</span>so the kinetic energy is equal to
=29.4 J
Answer:
v ’= - 1.76 10⁻⁴ m / s
Explanation:
We can solve this problem using momentum conservation. Defined a system formed by the patient, his blood and the platform in such a way that the forces are internal and the moment is conserved
initial instnate. Before pumping
p₀ = 0
final instant. Right after the heart pumping
p_f = m v + M v'
where m is the mass of blood and M the mass of the patient + the platform
p₀ = p_f
0 = m v + M v’
v ’= - ![\frac{m}{M} \ v](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7BM%7D%20%5C%20v)
let's calculate
v ’= -
0.30
v ’= - 1.76 10⁻⁴ m / s
To calculate the speed of the water leaving we use the formula,
Here h is the height of the water above the hole and g is the acceleration due to gravity.
Given,
Substituting the given value in above formula we get
Thus, water flow from a hole at the speed is 10.7 m/s
To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:
![\lambda = \frac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bv%7D%7Bf%7D)
Where,
Wavelength
f = Frequency
v = Velocity
Our values are given as,
![f = 2.8*10^3Hz](https://tex.z-dn.net/?f=f%20%3D%202.8%2A10%5E3Hz)
Speed of sound
Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:
Replacing we have,
![\lambda = \frac{340}{2.8*10^3}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B340%7D%7B2.8%2A10%5E3%7D)
![\lambda = 0.1214m](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.1214m)
Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m