________ delivers the precise number of parts to be assembled into a finished product at precisely the right time.

A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. a)

maks197457 [2]

Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars

Explanation :

A)

Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}

= 3780kJ

And 1 hour = 3600s

Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W

B)

At 15km/hour a 15km run takes 1 hour.

1 hour is 3600s and the runner burns 1050 joule per second.

Energy used in 1 hour = 3600 x 1050 J/s

= 3780000 J or 3.78MJ

C)

1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km

15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ

Finally,

1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ

This means that the runner needs 5320/1008 = 5.3 bars

7 0

3 years ago

Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.

atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage $v_p=120volt$

(i) We know that peak voltage is give by $v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt$

(ii) We know that for transformer $\frac{v_p}{v_s}=\frac{n_p}{n_s}$

So $\frac{169.08}{v_s}=\frac{10}{1}$

$v_s=16.90volt$

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by $v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt$

(v) Now dc current is given by $i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A$

4 0

2 years ago

What is a Wayfaring graphic?

KonstantinChe [14]

Answer:

ambages. pitiless or without compassion; cruel; merciless. winding, roundabout paths or ways.

Explanation:

6 0

3 years ago

Air fl ows isentropically through a duct. At section 1, the pressure and temperature are 250 kPa and 1258C, and the velocity is

abruzzese [7]

The correct temperature is 125°C

Answer:

A) M_a1 = 0.5

B) T2 = 232.17 K

C) V2 = 611 m/s

D) m' = 187 kg/s

Explanation:

We are given;

Pressure; P1 = 250 kPa

Temperature; T1 = 125°C = 398 K

Speed; v1 = 200 m/s

Area; A2 = 0.25 m²

M_a2 = 2

A) Formula for M_a1 is given by;

M_a1 = v/a1

Where;

v is speed

a1 = √kRT

k is specific heat capacity ratio of air = 1.4

R is a gas constant with a value of R = 287 J/kg·K

T is temperature

Thus;

M_a1 = 200/√(1.4 × 287 × 398)

M_a1 = 200/399.895

M_a1 = 0.5

B) To find T2, let's first find the Stagnation pressure T0

Thus;

T0/T1 = 1 + ((k - 1)/2) × (M_a1)²

T0 = T1(1 + ((k - 1)/2) × (M_a1)²)

T0 = 398(1 + ((1.4 - 1)/2) × (0.5)²)

T0 = 398(1 + (0.2 × 0.5²))

T0 = 398 × 1.05

T0 = 417.9 K

Now,similarly;

T0/T2 = 1 + ((k - 1)/2) × (M_a2)²

T2 = T0/[(1 + ((k - 1)/2) × (M_a2)²)]

T2 = 417.9/(1 + (0.2 × 2²))

T2 = 417.9/1.8

T2 = 232.17 K

C) V2 is gotten from the formula;

T0 = T2 + (V2)²/(2C_p)

Cp of air = 1005 J/Kg.K

Thus;

V2 = √(2C_p)[T0 - T2]

V2 = √((2 × 1005) × (417.9 - 232.17))

V2 =√373317.3

V2 = 611 m/s

D) mass flow is given by the formula;

m' = ρA2•V2

Where;

ρ is Density of air with an average value of 1.225 kg/m³

m' = 1.225 × 0.25 × 611

m' = 187 kg/s

6 0

2 years ago

WHICH COMPONENT IN A BLOWER MOTOR

bagirrra123 [75]

I think it would be the resistor?

5 0

2 years ago