Answer:
Explanation:
When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.
So torque exerted by wood on the blade
= force x perpendicular distance from the axis of rotation
= 104 x .128
=13.312 Nm.
Since this torque opposes the movement of blade , it turns the blade slower.
Answer:
A) 60%
B) p2 = 1237.2 kPa
v2 = 0.348 m^3
C) w1-2 = w3-4 = 1615.5 kJ
Q2-3 = 60 kJ
Explanation:
A) calculate thermal efficiency
Л = 1 -
where Tl = 300 k
Th = 750 k
hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%
B) calculate the pressure and volume at the beginning of the isothermal expansion
calculate pressure ( P2 ) :
= P3v3 = mRT3 ----- (1)
v3 = 0.4m , mR = 2* 0.287, T3 = 750
hence P3 = 1076.25
next equation to determine P2
Qex = p3v3 ln( p2/p3 )
60 = 1076.25 * 0.4 ln(p2/p3)
hence ; P2 = 1237.2 kpa
calculate volume ( V2 )
p2v2 = p3v3
v2 = p3v3 / p2
= (1076.25 * 0.4 ) / 1237.2
= 0.348 m^3
C) calculate the work and heat transfer for each four processes
work :
W1-2 = mCv( T2 - T1 )
= 2*0.718 ( 750 - 300 ) = 1615.5 kJ
W3-4 = 1615.5 kJ
heat transfer
Q2-3 = W2-3 = 60KJ
Q3-4 = 0
D ) sketch of the cycle on p-V coordinates
attached below
This room is called Substation
Answer:
![r_{cm}=[12.73,12.73]cm](https://tex.z-dn.net/?f=r_%7Bcm%7D%3D%5B12.73%2C12.73%5Dcm)
Explanation:
The general equation to calculate the center of mass is:
![r_{cm}=1/M*\int\limits {r} \, dm](https://tex.z-dn.net/?f=r_%7Bcm%7D%3D1%2FM%2A%5Cint%5Climits%20%7Br%7D%20%5C%2C%20dm)
Any differential of mass can be calculated as:
Where "a" is the radius of the circle and λ is the linear density of the wire.
The linear density is given by:
![\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}](https://tex.z-dn.net/?f=%5Clambda%3DM%2FL%3DM%2F%28a%2A%5Cpi%2F2%29%3D%5Cfrac%7B2M%7D%7Ba%5Cpi%7D)
So, the differential of mass is:
![dm = \frac{2M}{a\pi}*a*d\theta](https://tex.z-dn.net/?f=dm%20%3D%20%5Cfrac%7B2M%7D%7Ba%5Cpi%7D%2Aa%2Ad%5Ctheta)
![dm = \frac{2M}{\pi}*d\theta](https://tex.z-dn.net/?f=dm%20%3D%20%5Cfrac%7B2M%7D%7B%5Cpi%7D%2Ad%5Ctheta)
Now we proceed to calculate X and Y coordinates of the center of mass separately:
![X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta](https://tex.z-dn.net/?f=X_%7Bcm%7D%3D1%2FM%2A%5Cint%5Climits%5E%7B%5Cpi%2F2%7D_0%20%7Ba%2Acos%5Ctheta%2A2M%2F%5Cpi%7D%20%5C%2C%20d%5Ctheta)
![Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta](https://tex.z-dn.net/?f=Y_%7Bcm%7D%3D1%2FM%2A%5Cint%5Climits%5E%7B%5Cpi%2F2%7D_0%20%7Ba%2Asin%5Ctheta%2A2M%2F%5Cpi%7D%20%5C%2C%20d%5Ctheta)
Solving both integrals, we get:
![X_{cm}=2*a/\pi=12.73cm](https://tex.z-dn.net/?f=X_%7Bcm%7D%3D2%2Aa%2F%5Cpi%3D12.73cm)
![Y_{cm}=2*a/\pi=12.73cm](https://tex.z-dn.net/?f=Y_%7Bcm%7D%3D2%2Aa%2F%5Cpi%3D12.73cm)
Therefore, the position of the center of mass is:
![r_{cm}=[12.73,12.73]cm](https://tex.z-dn.net/?f=r_%7Bcm%7D%3D%5B12.73%2C12.73%5Dcm)
Air pressure is the weight of air on an area. The weight of air
is due to the gravitational forces between the Earth and the
molecules of its atmosphere.