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nirvana33 [79]
3 years ago
14

Which equation can be used to find x, the length of the hypotenuse of the right triangle? A triangle has side lengths 63, 16, x.

16 + 63 = x 16 squared + 63 = x (16 + 63) squared = x squared 16 squared + 63 squared = x squared
Engineering
2 answers:
SIZIF [17.4K]3 years ago
8 0

Answer:

16 squared + 63 squared = x squared

Explanation:

Kryger [21]3 years ago
3 0

Answer: 16 squared + 63 squared = x squared

Explanation:

Hi, since we have a right triangle we have to apply the Pythagorean Theorem:  

c^2 = a^2 + b^2  

Where c is the hypotenuse of the triangle (the longest side) and a and b are the other sides.  

Replacing with the values given:  

x^2 = 63^2 + 16^2  

So, the correct option is  

16 squared + 63 squared = x squared

Feel free to ask for more if needed or if you did not understand something.  

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A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa
taurus [48]

Answer:

Explanation:

Given

T_h=250^{\circ}C\approx 523\ K

T_L=30^{\circ}C\approx 303\ K

Q_1=6 kW

From Clausius inequality

\oint \frac{dQ}{T}=0  =Reversible cycle

\oint \frac{dQ}{T}  =Irreversible cycle

\oint \frac{dQ}{T}>0  =Impossible

(a)For P_{out}=3 kW

Rejected heat Q_2=6-3=3\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K

thus it is Impossible cycle

(b)P_{out}=2 kW

Q_2=6-2=4 kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K

Possible

(c)Carnot cycle

\frac{Q_2}{Q_1}=\frac{T_1}{T_2}

Q_2=3.47\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3.47}{303}=0

and maximum Work is obtained for reversible cycle when operate between same temperature limits

P_{out}=Q_1-Q_2=6-3.47=2.53\ kW

Thus it is possible

6 0
4 years ago
A three-phase voltage source with a terminal voltage of 22kV is connected to a three-phase transformer rated 5MVA 22kV/220V. The
amid [387]

Answer:

A) See Attachment B) See Attachment C) 186.153

Explanation:

C) Per phase Voltage= 220/∛3

                                  = 6.037

S=V²/Z

  = (6.037)²/(0.01+i0.05)

  = 62051.282-i310256.41

Active power losses per phase= 62.051kW

total Active power losses= 62.051×3

                                         =186.153kW

6 0
4 years ago
The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for tim
AysviL [449]

Answer:

A.) 1mv = 2000N

B.) Impulse = 60Ns

C.) Acceleration = 66.67 m/s^2

Velocity = 4 m/s

Displacement = 0.075 metre

Absorbed energy = 60 J

Explanation:

A.) Using a mathematical linear equation,

Y = MX + C

Where M = (2000 - 0)/( 898 - 0 )

M = 2000/898

M = 2.23

Let Y = 2000 and X = 898

2000 = 2.23(898) + C

2000 = 2000 + C

C = 0

We can therefore conclude that

1 mV = 2000N

B.) Impulse is the product of force and time.

Also, impulse = momentum

Given that

Mass M = 30kg

Velocity V = 2 m/s

Impulse = M × V = momentum

Impulse = 30 × 2 = 60 Ns

C.) Force = mass × acceleration

F = ma

Substitute force and mass into the formula

2000 = 30a

Make a the subject of formula

a = 2000/30

acceleration a = 66.67 m/s^2

Since impulse = 60 Ns

From Newton 2nd law,

Force = rate of change in momentum

Where

change in momentum = -MV - (- MU)

Impulse = -MV + MU

Where U = initial velocity

60 = -60 + MU

30U = 120

U = 120/30

U = 4 m/s

Force = 2000N

Impulse = Ft

Substitute force and impulse to get time

60 = 2000t

t = 60/2000

t = 0.03 second

Using third equation of motion

V^2 = U^2 + 2as

Where S = displacement

4^2 = 2^2 + 2 × 66.67S

16 = 4 + 133.4S

133.4S = 10

S = 10/133.4

S = 0.075 metre

D.) Energy = 1/2 mV^2

Energy = 0.5 × 30 × 2^2

Energy = 15 × 4 = 60J

5 0
3 years ago
Fluids at rest possess no flow energy. a)- True b)- False
gtnhenbr [62]

Answer:

True.

Explanation:

According the engineering flow they don not possess flow energy when they are in rest.

When they are in motion they show a translation energy.

The features if fluids may be different according the variables of pressure and temperature.

8 0
3 years ago
You have a 12-inch PVC water main that is 850 feet long flowing at 5.6 cfs. Point A is at an elevation of 750 ft. Point B is at
alex41 [277]

Known :

D = 12 in = 1 ft

L = 850 ft

Q = 5.6 cfs

hA = 750 ft

hB = 765 ft

PA = 85 psi = 12240 lb/ft²

Solution :

A = πD² / 4 = π(1²) / 4

A = 0.785 ft²

<u>Velocity of water :</u>

U = Q / A = 5.6 / 0.785

U = 7.134 ft/s

<u>Friction loss due to pipe length :</u>

Re = UD / v = (7.134)(1) / (0.511 × 10^(-5))

Re = 1.4 × 10⁶

(From Moody Chart, We Get f = 0.015)

hf = f(L / d)(U² / 2g) = 0.015(850 / 1)((7.134²) / 2(32.2))

hf = 10 ft

PA + γhA = PB + γhB + γhf

PB = PA + γ(hA - hB - hf)

PB = 12240 + (62.4)(750 - 765 - 10)

PB = 10680 lb/ft²

PB = 74.167 psi

8 0
3 years ago
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