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Fiesta28 [93]
2 years ago
12

How does warming up the tires on a car increase grip with the pavement?

Engineering
1 answer:
ICE Princess25 [194]2 years ago
8 0

Answer:

because burning rubber increases the grip power

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Rolling and Shearing are the types of a)-Bulk Deformation Process b)- Sheet Metal Process c)- Machining Process d)- Both a &
Margarita [4]

Answer:

a)Bulk deformation process  

Explanation:

<u>Rolling</u>

Rolling is a metal forming process.In rolling work piece passes through two moving rollers and get compressed.in rolling thickness of work piece will reduces and length of work piece will increase for maintaining the constant area.Due to compression bulk deformation takes place.

<u>Shearing</u>

In shearing one surface slides on another surface and deformation take place.shearing is a machining process.This is also a bilk motion deformation process.

So from above we can say that option a is right.

5 0
3 years ago
The housing for a certain machinery product is made of two components, both aluminum castings. The larger component has the shap
Dmitry_Shevchenko [17]

Answer:

The pouring of the molten metal during casting is done very slowly hence the molten metal froze before reaching all parts of the mould cavity. also the early freezing can occur if the temperature of the molten metal was lower than the required temperature for casting

Explanation:

since the same components are being casted at other foundries and they don't have the defects ,

Hence the reason for the defects experienced by these components can be caused when the pouring of the molten metal during casting is done very slowly hence the molten metal froze before reaching all parts of the mould cavity. also the early freezing can occur if the temperature of the molten metal was lower than the required temperature for casting .

8 0
3 years ago
The Web and Digital Communications pathway is broken down into four main categories of technology. Which category BEST describes
jasenka [17]
D marketing because he’s advertising ski equipment to buy
6 0
2 years ago
Pls help me with these 3 ez questions.
Katyanochek1 [597]
15- T
16- F
17- T
I think this is correct
7 0
2 years ago
Read 2 more answers
A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}

  r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Now by putting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

  Ф=18.03°

So the shear angle is 18.03°.

4 0
3 years ago
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