Why did you put this on here when you know the answer lol
Answer:
The provided length of the vertical curve is satisfactory for the reconstruction design speed of 60 mi/h
Explanation:
The explanation is shown on the first uploaded image
Answer:
The spring is compressed by 0.275 meters.
Explanation:
For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston
Mathematically we can write
![Force_{pressure}=Force_{spring}+Weight_{piston}](https://tex.z-dn.net/?f=Force_%7Bpressure%7D%3DForce_%7Bspring%7D%2BWeight_%7Bpiston%7D)
we know that
![Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons](https://tex.z-dn.net/?f=Force_%7Bpressure%7D%3DPressure%5Ctimes%20Area%3D300%5Ctimes%2010%5E%7B3%7D%5Ctimes%20%5Cfrac%7B%5Cpi%20%5Ctimes%200.1%5E2%7D%7B4%7D%3D750%5Cpi%20Newtons)
![Weight_{piston}=mass\times g=100\times 9.81=981Newtons](https://tex.z-dn.net/?f=Weight_%7Bpiston%7D%3Dmass%5Ctimes%20g%3D100%5Ctimes%209.81%3D981Newtons)
Now the force exerted by an spring compressed by a distance 'x' is given by ![Force_{spring}=k\cdot x=5\times 10^{3}\times x](https://tex.z-dn.net/?f=Force_%7Bspring%7D%3Dk%5Ccdot%20x%3D5%5Ctimes%2010%5E%7B3%7D%5Ctimes%20x)
Using the above quatities in the above relation we get
![5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E%7B3%7D%5Ctimes%20x%2B981%3D750%5Cpi%20%5C%5C%5C%5C%5Ctherefore%20x%3D%5Cfrac%7B750%5Cpi%20-981%7D%7B5%5Ctimes%2010%5E%7B3%7D%7D%3D0.275meters)
Answer:
The best saw for cutting miter joints is the backsaw.
Add-on:
i hope this helped at all.
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ