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Volgvan
3 years ago
10

What is a transition? A. An animation that happens on a single slide B. An outline format that uses roman numerals C. An image f

ile imported to a title slide D. An effect that happens between slides Please select the best answer from the choices provided A B C D
Engineering
1 answer:
True [87]3 years ago
4 0

Answer:

its c

Explanation:

You might be interested in
A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower
PIT_PIT [208]

Answer:

γ_{xy} =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ_{xy} = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ_{xy}

By comapring both equations, we get

P/A = Gγ_{xy}   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0
3 years ago
Answer every question of this quiz
Reil [10]

I'd say number 4, number 3 looks like an exhaust valve

5 0
4 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
AURORKA [14]

Answer:

a) w_{out} = 281.55\,\frac{kJ}{kg}, b) s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

-q_{out} - w_{out} + h_{in}-h_{out} = 0

w_{out} = h_{in} - h_{out}-q_{out}

w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}

w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}

w_{out} = 281.55\,\frac{kJ}{kg}

b) The entropy production is determined after the Second Law of Thermodynamics:

-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0

s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}

s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)

s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)

s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}

3 0
3 years ago
Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Co
butalik [34]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

5 0
3 years ago
a hallow steel tube 3.5m long has external diameter of 120mm. in order to determine the internal diameter the tube was subjected
Dafna1 [17]

Answer:

Please check the photo (solve)

4 0
3 years ago
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