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ASHA 777 [7]
3 years ago
6

What are the advantages to a quality saw?

Engineering
1 answer:
gladu [14]3 years ago
7 0
Ans: A quality handsaw makes jobs such as crosscutting moldings, cleaning out dovetails and flush cutting dowels and pegs faster, cleaner and safer. There are myriad different types and sizes of blades for working with wood and other materials.
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Some cars have an FCW, which stands for
Natalka [10]

Answer:

FCW in car stands for <em>Forward Collision Warning. </em>

<u>Explanation:</u>

The vehicle speed is monitored by <em>FCW system</em>, this is an advanced technology which indicates to the rear vehicle that a crash is going to happen if the vehicle gets close <em>because of speed</em>. This FCW systems monitor’s distance between the vehicles and speed of the vehicles.

<em>FCW system do not control the vehicle completely</em>. This system consists of sensors to detect stationary or slower-moving vehicles. A signal alerts the driver if the <em>distance between the vehicles is less</em> so that crash is being happened. It helps driver from crash by changing his route. Cars with this technology consists of audible alert.  

8 0
3 years ago
a vertical cylindrical container is being cooled in ambient air at 25 °C with no air circulation. if the initial temperature of
Sloan [31]

Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period.  = 4.93 w/m²k

Explanation:

check attachement for answer explanation

7 0
3 years ago
Read 2 more answers
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
4 years ago
Who can help me with electric systems for cars?
hoa [83]

Answer: i can see if i can what is the problem

Explanation:

7 0
3 years ago
If they opened up the International Space Station to tourism, would you go? Why? answer in 2 sentences
Arlecino [84]
Personally, I would go to the space station. The space station has extreme different levels of technology and abilities, plus who doesn’t want to go to space.
6 0
3 years ago
Read 2 more answers
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