Can <span>get your heart rate up and increases blood circulation throughout the body.</span>
Answer:
16.8 lb is the force on the brake pad of one wheel.
Explanation:
Force applied on the piston = ![F_1=5.6 lb](https://tex.z-dn.net/?f=F_1%3D5.6%20lb)
Area of the piston = ![A_1=0.6 inches^2](https://tex.z-dn.net/?f=A_1%3D0.6%20inches%5E2)
Force applied on the brakes = ![F_2](https://tex.z-dn.net/?f=F_2)
Area of the brakes = ![A_2=1.8 inches^2](https://tex.z-dn.net/?f=A_2%3D1.8%20inches%5E2)
Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.
![\frac{F_1}{A_2}=\frac{F_2}{A_2}](https://tex.z-dn.net/?f=%5Cfrac%7BF_1%7D%7BA_2%7D%3D%5Cfrac%7BF_2%7D%7BA_2%7D)
![F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb](https://tex.z-dn.net/?f=F_2%3D%5Cfrac%7B5.6%20lb%5Ctimes%201.8%20inhes%5E2%7D%7B0.6%20inches%5E2%7D%3D16.8%20lb)
16.8 lb is the force on the brake pad of one wheel.
Answer:
The workdone is ![W = 9.28 * 10^{3} J](https://tex.z-dn.net/?f=W%20%3D%209.28%20%2A%2010%5E%7B3%7D%20J)
Explanation:
From the question we are told that
The height of the cylinder is ![h = 0.588\ m](https://tex.z-dn.net/?f=h%20%3D%200.588%5C%20m)
The face Area is ![A = 4.19 \ m^2](https://tex.z-dn.net/?f=A%20%3D%204.19%20%5C%20m%5E2)
The density of the cylinder is ![\rho = 0.346 * \rho_w](https://tex.z-dn.net/?f=%5Crho%20%20%3D%20%200.346%20%2A%20%5Crho_w)
Where
is the density of freshwater which has a constant value
![\rho_w = 1000 kg/m^3](https://tex.z-dn.net/?f=%5Crho_w%20%3D%201000%20kg%2Fm%5E3)
Now
Let the final height of the device under the water be ![= h_f](https://tex.z-dn.net/?f=%3D%20%20h_f)
Let the initial volume underwater be ![= V_n](https://tex.z-dn.net/?f=%3D%20V_n)
Let the initial height under water be ![= h_i](https://tex.z-dn.net/?f=%3D%20h_i)
Let the final volume under water be ![= V_f](https://tex.z-dn.net/?f=%3D%20V_f)
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as
![\rho_c g V_n = \rho_w gV_f](https://tex.z-dn.net/?f=%5Crho_c%20g%20V_n%20%3D%20%5Crho_w%20gV_f)
![\rho_c A h = \rho A h_f](https://tex.z-dn.net/?f=%5Crho_c%20A%20h%20%3D%20%5Crho%20A%20h_f)
So ![\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}](https://tex.z-dn.net/?f=%5Cfrac%7B0.346%20%5Crho_w%7D%7B%5Crho_w%7D%20%3D%20%5Cfrac%7Bh_f%7D%7Bh%7D)
=> ![\frac{h_f}{h_c} = 0.346](https://tex.z-dn.net/?f=%5Cfrac%7Bh_f%7D%7Bh_c%7D%20%20%3D%200.346)
Now the work done is mathematically represented as
![W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5E%7Bh_f%7D_%7Bh%7D%20%7B%5Crho_w%20g%20A%20%28-h%29%7D%20%5C%2C%20dh)
![= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.](https://tex.z-dn.net/?f=%3D%20%20%20%5Crho_w%20g%20A%20%5B%5Cfrac%7Bh%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20h_f%7D%20%5Catop%20%7Bh%7D%7D%20%5Cright.)
![= \frac{g A \rho}{2} [h^2 - h_f^2]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%20%5Bh%5E2%20-%20h_f%5E2%5D)
![= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%28h%5E2%29%20%20%5B1%20%20-%20%5Cfrac%7Bh_f%5E2%7D%7Bh%5E2%7D%20%5D)
Substituting values
![W = 9.28 * 10^{3} J](https://tex.z-dn.net/?f=W%20%3D%209.28%20%2A%2010%5E%7B3%7D%20J)
Answer:
![E=8.02*10^{5}J](https://tex.z-dn.net/?f=E%3D8.02%2A10%5E%7B5%7DJ)
Explanation:
Given data
Electromagnetic waves from the sun is I=1.4kW/m² at 80%
Area a=(0.30×0.51)m²
Time t=1.30 hr
To find
Energy E
Solution
The energy received by your back is calculated as:
![E=Pt=Iat\\E=(0.80)(1400W/m^{2} )(0.30*0.51m^{2} )(1.30*3600s)\\E=8.02*10^{5}J](https://tex.z-dn.net/?f=E%3DPt%3DIat%5C%5CE%3D%280.80%29%281400W%2Fm%5E%7B2%7D%20%29%280.30%2A0.51m%5E%7B2%7D%20%29%281.30%2A3600s%29%5C%5CE%3D8.02%2A10%5E%7B5%7DJ)