Answer:
The required frequency = 0.442 Hz
Explanation:
Frequency ![f = ( \dfrac{1}{2 \pi}) \omega](https://tex.z-dn.net/?f=f%20%20%3D%20%28%20%5Cdfrac%7B1%7D%7B2%20%5Cpi%7D%29%20%5Comega)
where;
![\omega = \sqrt{\dfrac{k}{m} }](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%20%7D)
Then;
![f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )](https://tex.z-dn.net/?f=f%20%3D%20%5CBigg%20%28%20%5Cdfrac%7B1%7D%7B2%20%5Cpi%7D%20%20%5CBigg%20%29%20%20%20%5CBigg%28%20%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%20%7D%20%20%5CBigg%20%29)
However;
and;
mass ![m = m_{car } + m_{person}](https://tex.z-dn.net/?f=m%20%3D%20m_%7Bcar%20%7D%20%2B%20m_%7Bperson%7D)
![f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )](https://tex.z-dn.net/?f=f%20%3D%20%5CBigg%20%28%20%5Cdfrac%7B1%7D%7B2%20%5Cpi%7D%20%20%5CBigg%20%29%20%20%20%5CBigg%28%20%5Csqrt%7B%5Cdfrac%7B%5Cdfrac%7BF%7D%7Bx%7D%7D%7Bm_%7Bcar%7D%2Bm_%7Bperson%7D%7D%20%7D%20%20%5CBigg%20%29)
![f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )](https://tex.z-dn.net/?f=f%20%3D%20%5CBigg%20%28%20%5Cdfrac%7B1%7D%7B2%20%5Cpi%7D%20%20%5CBigg%20%29%20%20%20%5CBigg%28%20%5Csqrt%7B%5Cdfrac%7B%7BF%7D%7D%7Bx%28m_%7Bcar%7D%2Bm_%7Bperson%7D%29%7D%20%7D%20%20%5CBigg%20%29)
where;
![F = m_{person}g](https://tex.z-dn.net/?f=F%20%3D%20m_%7Bperson%7Dg)
Then;
![f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )](https://tex.z-dn.net/?f=f%20%3D%20%5CBigg%20%28%20%5Cdfrac%7B1%7D%7B2%20%5Cpi%7D%20%20%5CBigg%20%29%20%20%20%5CBigg%28%20%5Csqrt%7B%5Cdfrac%7B%20%7Bm_%7Bperson%7Dg%20%7D%7D%7Bx%28m_%7Bcar%7D%2Bm_%7Bperson%7D%29%7D%20%7D%20%20%5CBigg%20%29)
replacing the values;
![f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )](https://tex.z-dn.net/?f=f%20%3D%20%5CBigg%20%28%20%5Cdfrac%7B1%7D%7B2%20%5Cpi%7D%20%20%5CBigg%20%29%20%20%20%5CBigg%28%20%5Csqrt%7B%5Cdfrac%7B%20%7B%2888%20%5C%20kg%29%2A%20%289.81%20%5C%20m%2Fs%5E2%29%20%7D%7D%7B%285.36%20%5Ctimes%2010%5E%7B-2%7D%20%5C%20m%29%20%282002%20%5C%20kg%20%2B88%20%5C%20kg%29%7D%20%7D%20%20%5CBigg%20%29)
![\mathbf{f = 0.442 \ Hz}](https://tex.z-dn.net/?f=%5Cmathbf%7Bf%20%3D%200.442%20%5C%20Hz%7D)
Answer:
Subtract the kinetic energy at the bottom from the potential energy loss. The remainder becomes frictional heat.
Potential energy loss:
M g H = 21.7*9.81*3.5 = 745.1 J
Kinetic energy at bottom of slide:
= (1/2) M v^2 = 52.5 J
Answer:
34000 J or 34 kJ
Explanation:
From the question, The energy needed to just melt the ice cube does not requires any external temperature or change in temperature, Hence it is called latent heat
By Applying,
Q' = C'm........................ Equation 1
Where Q' = Latent heat, C' = Specific latent heat of fusion of water, m = mass of ice.
Given: C' = 340000 J/kg, m = 100 g = 0.1 kg
Substitute into equation 1
Q' = 0.1(340000)
Q' = 34000 J
Q' = 34 kJ
The answer for this question is negative externality