ENGINEERING COLLEGE

Convert the angles of a triangle to radians.Part A31∘43′53′′, 90∘32′11′′, 57∘43′56′′Express your answers, separated by commas, to six significant figures.nothingrad, rad, radRequest AnswerPart B94∘22′19′′, 40∘54′53′′, 44∘42′48′′Express your answers, separated by commas, to six significant figures.

Answers

Answer 1

Answer:

Answer:

Explanation:

To convert to radians

A31∘43′53′′, 90∘32′11′′, 57∘43′56′′

using DMS approach ; 1degree = 60minutes = 3600 seconds

1° = 60' = 3600"

And degree to radian = multiply by π/180

A) 31∘43′53′′ = 31degree + 43minutes + 53 seconds

= 31 degree + 43minutes + 53/60

= 31 degree + 43.88minutes

= 31 degree + 43.88/60 = 31.73 degree x π/180 = 0.5534radians

FOR 90∘32′11′′ = 90 degree + 32minutes + 11seconds

= 90degree + 32minutes + 11/60

= 90 degree + 32.183minutes

= 90 degree + 32.183/60 = 90.54degree x π/180

= 1.580radians

FOR 57∘43′56′′ = 57degree + 43minutes+ 56seconds

= 57degree + 43minutes + 56/60

= 57 degree + 43.93minutes

= 57degree + 43.93/60 = 57.73degree X π/180

= 1.00radians

PART B

FOR 94∘22′19′′ = 94degree + 22minutes + 19seconds

= 94degree + 22minutes + 19/60

= 94degree + 22.32minutes

= 94degree + 22.32/60

= 94.37degree X π/180 = 1.65radians

FOR 40∘54′53′′ = 40degree + 54minutes + 53seconds

= 40 degree + 54minutes + 53/60

= 40 degree + 54.88minutes = 40 degree + 54.88/60

= 40.91degree X π/180 = 0.714radians

FOR 44∘42′48′′ = 44degree + 42minutes + 48seconds

= 44degree + 42.8minutes

= 44.71degree X π/180 = 0.780radians

Answer 2

Answer:

0.176270π rad, 0.502980π rad, 0.320735π rad

0.524289π rad, 0.227304π rad, 0.248407π rad

Explanation:

We know that,

1° = 60' 180° = π

1 ' = 1°/60 1° = π/180

a. 31°43'53''

Step 1

53'' = 53 * 1/60

= 53'/60

Step 2

43'53''

= 43'+53'/60

= (2580+43)/60

= 2623'/60

-------- Convert to degrees

= 2623/60 * 1/60

= 2623/3600

Step 3

31°43'53''

= 31+ 2623/3600

= (111600 + 2623)/3600

= 114223°/3600

Now, we convert to radians

= 114223/3600 * π/180°

= 0.176270π rad

90°32'11''

Step 1.

11' = 11 * 1/60

= 11/60

Step 2

32'11'

= 32 + 11/60

= 1931/60

-------- Convert to degrees

= 1931/60 * 1/60

= 1931/3600

Step 3

90°31'11''

= 90 + 1931/3600

= 325931°/3600

Now we convert to radians

= 325931°/3600 * π/180°

= 0.502980π rad

57°43'56''

Step 1

56' = 56 * 1/60

= 56/60

= 14/15

Step 2

43'56''

= 43 + 14/15

= 659/15

Now we convert to degrees

= 659/15 * 1/60

= 659°/900

Step 3

57°43'56''

= 57 + 659/900

= 51959/900

Now we convert to radians

= 51959°/900 * π/180°

= 0.320735π rad

94∘22′19′′

Step 1

19'' = 19/60

Step 2

22'19''

= 22 + 19/60

= 1339/60

Now we convert to degrees

= 1339/60 * 1/60

= 1339°/3600

Step 3

94°22'19"

= 94 + 1339/3600

= 339739°/3600

Now we convert to radians

= 339739°/3600 * π/180

= 0.524289π rad

40∘54′53′′

Step 1

53" = 53/60

Step 2

54'53"

= 54'+ 53/60

= 3293/60

Now we convert to degrees

= 3293/60 * 1/60

= 3293/3600

Step 3

40°54'53"

= 40 + 3293/3600

= 147293/3600

Now we convert to radians

= 147293/3600 * π/180

= 0.227304π rad

44∘42′48′

Step 1

48' = 48/69

= 4/5

Step 2

42'48"

= 42 + 4/5

=214/5

Nowz we convert to degrees

= 214/5 * 1/60

= 107/150

Step 3

44°42'48"

= 44 + 107/150

= 6707/150

Now we convert to radians

= 6707/150 * π/180

= 0.248407π rad

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A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C?

Answers

Answer:

$\eta _(max) = 0.2413 = 24.13%$

$\eta' _(max) = 0.5061 = 50.61%$

Given:

$T_(1max) = 100^(\circ) = 273 + 100 = 373 K$

operating temperature of heat engine, $T_(2) = 10^(\circ) = 273 + 10 = 283 K$

$T_(3max) = 300^(\circ) = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _(max)$ is given by:

$\eta _(max) = 1 - (T_(2))/(T_(1max))$

$\eta _(max) = 1 - (283)/(373) = 0.24$

$\eta _(max) = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_(3max)$ changes to $300^(\circ)$ , so, the new maximum efficiency, $\eta' _(max)$ is given by:

$\eta' _(max) = 1 - (T_(2))/(T_(3max))$

$\eta _(max) = 1 - (283)/(573) = 0.5061$

$\eta _(max) = 0.5061 = 50.61%$

The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal

Answers

Answer:

Isobaric process

Explanation:

The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.

At constant pressure, the work done is given by :

$W=p* \Delta V$

Where

W is the work done by the system

p is the constant pressure

$\Delta V$ is the change in volume

So, the correct option is (c) " isobaric process ".

In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetnic

Answers

Answer:

The correct answer is C) Trimetric

Explanation:

The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answers

Answer:

$v = 250[1 - {e^(-6000t)}]$ mV

Explanation:

The voltage across a capacitor at a time t, is given by:

$v(t) = (1)/(C) \int\limits^(t)_(t_0) {i(t)} \, dt + v(t_0)$ ----------------(i)

Where;

v(t) = voltage at time t

$t_(0)$ = initial time

C = capacitance of the capacitor

i(t) = current through the capacitor at time t

v(t₀) = voltage at initial time.

From the question:

C = 2μF = 2 x 10⁻⁶F

i(t) = 3 $e^(-6000t)$ mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + v(0)$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + 0$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt$

$v = (3)/(2*10^(-6)) \int\limits^(t)_(0) {e^(-6000t)} \, dt$ [Solve the integral]

$v = (3)/(2*10^(-6)*(-6000)) {e^(-6000t)}|_0^t$

$v = (-3000)/(12) {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)} - [-250 {e^(-6000(0))]$

$v = -250 {e^(-6000t)} - [-250]$

$v = -250 {e^(-6000t)} + 250$

$v = 250 -250 {e^(-6000t)}$

$v = 250[1 - {e^(-6000t)}]$

Therefore, the voltage across the capacitor is $v = 250[1 - {e^(-6000t)}]$ mV

Determine the degrees of superheat of steam at 101.325 kPa and 170°C. a. 50°C b. 70°C c. 60°C d. 80°C

Answers

Answer:

b) 70°C

Explanation:

Given that super heat temperature at 101.325 KPa= 170°C.

We know that saturation temperature at 101.325 KPa is 100°C.

Degree of super heat =Super heat temperature - saturation temperature ( at constant pressure)

Now by putting the values

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Degree of super heat=70°C.

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Answers

Answer:

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Explanation:

Other than the scientific reasons listed in the question, one of the main reasons why people all over the world are pursuing this endeavor is exploration. As human beings, we love to imagine new worlds and life-forms that we have never seen before. This fuels our need for exploration. Scientists throughout generations have dedicated their entire lives to learning and creating newer and better technology in order for humans to take that next step in exploring and learning the secrets of the universe.

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