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3 years ago

Cutting and forming operations performed on relatively thin sheets of metal. a)- True b)-False

Engineering

1 answer:

belka [17]3 years ago

6 0

Answer: True

Explanation: There are different actions that are performed on the metal sheet . Cutting operation is the operation of splitting,slitting or parting of the metal sheet on the by using different kind of blades . Formation operation are basically for the exerting pressure on the sheet to give it desired shape such as bending, squeezing etc. Therefore the given statement is true.

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Three 12-V, 100-A-hr batteries are connected in series. What are the output voltage and A-hr capacity of this connection

Lera25 [3.4K]

Answer:

36V, 100A-hr

Explanation:

Since the batteries are connected in series;

i. the output voltage will be the sum of the individual voltages

ii. the current rating (A-hr) will be the same as their individual current rating (A-hr). And this is because the same current flows through the batteries.

From i,

The output voltage, V, is given by the sum of the voltages of the three batteries;

V = 12V + 12V + 12V

V = 36V

From ii,

The A-hr capacity of the connection is the same as that of the individual batteries;

100 A-hr

Therefore, the output voltage and A-hr capacity of this connection is:

36V, 100A-hr

3 0

4 years ago

A 4 cm diameter sphere of copper is initially at a temperature of 95 °C. It is placed in a very large water bath at time t- 0. T

avanturin [10]

Answer:112.376 s

Explanation:

Given

$T_i=95^{\circ}C$

$T_f=35^{\circ}C$

$T_\infty \left ( ambient\right )=25^{\circ}C$

$h=400 watts/\left ( m^{2}^{\circ}C\right )$

$c=0.385 J/\left ( m^2^{\circ}C\right )$

$\rho =9 gm/cm^{3}$

Using Newton's law of cooling

$\frac{T_i-T_{\infty}}{T-T_{\infty}}$ = $e^{\frac{ht}{\rho L_{c}c}}$

$\frac{95-25}{35-25}$ = $e^{\frac{400\times 3\times 10^{-4}\times t}{9\times 2\times 0.385}}$

7= $e^{1.7316\times 10^{-2}\times t}$

Taking log both side

t=112.376sec

4 0

3 years ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

6 0

3 years ago

A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a

statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre = 183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0

3 years ago

University administrators have developed a Markov model to simulate graduation rates at their school. Students might drop out, r

abruzzese [7]

Solution :

The percentage of the students who have a chance of repeating their current year = 3%

The drop out students for the first year and the sophomores = 6%

Drop out rate of first year and the seniors = 4%

Now for the state space :

S = { first year(1), sophomores(2), juniors(3), seniors(4), graduates(G), Dropouts(D) }

Therefore

the first year students are indicated as '1'

Sophomores are indicated as '2'

Juniors are indicated as '3'

Seniors are indicated as '4

Graduates are indicated as 'G'

Dropouts are indicated as 'D'

The transition diagram is attached below.

The probability of the students who have the chance of repeating their current year = 3/100 = 0.03

Probability of first year dropouts and sophomores = 6/100 = 0.06

Probability of dropout rate of juniors and seniors = 4/100 = 0.04

Therefore, the probability matrix can be made as :

1 2 3 4 G D

$\begin{matrix}1\\ 2\\ 3\\ 4\\ G\\ D\end{matrix}$ $\begin{bmatrix}0.03 & 0.91 & 0 & 0 & 0 & 0.06\\ 0& 0.03 & 0.91 & 0 & 0 & 0.06\\ 0& 0 & 0.03 & 0.93 & 0 & 0.04\\ 0& 0 & 0 & 0.03 & 0.93 & 0.04\\ 0& 0 & 0 & 0 & 1 & 0\\ 0& 0 & 0 & 0 & 0 & 1\end{bmatrix}$

Here, G represents 'graduates' and D represents 'Dropouts.'

5 0

3 years ago