A stone-filled pit used for waste disposal is commonly referred to as a

A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of

Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

$E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%$

4 0

2 years ago

(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5

bezimeni [28]

Answer:

a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, $Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\$

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, $Q = \frac{109*0.5*100}{0.0075} = 726,666.667W$

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, $Q=\frac{1*0.5*100}{0.0075} = 6,666.67W$

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, $Q=\frac{109*0.5*100}{0.015} =363,333.33W$

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0

2 years ago

What tool should be used to loosen or tighten brake or fuel lines?

emmainna [20.7K]

The tool you would use are brake line wrenches.

4 0

3 years ago

Una empresa industrial agrupó sus fábricas de acuerdo con el valor de la producción anual en millones de C$ de cada una; se obtu

bulgar [2K]

Answer:

Explanation:

An industrial company grouped its factories according to the value of annual production in millions of C $ of each; the following distribution was obtained ...... Production value: 41-45, 46-50, 51-55, 56-60, 61-65, 66-70 ....... Number of factories: 7 , 10, 11, 9, 8, 7 ..... Complete the frequency distribution and determine and interpret the average annual production of the factories and the standard deviation of the productions.

5 0

2 years ago

Read 2 more answers

You are recording a friend's 3-minute song with 24 bits per sample at 96 kHz sampling rate for a 5.1 surround sound system (6 ch

djyliett [7]

Answer:

save space = 1.9626 Gibibits

Explanation:

given data

recording song time = 3 minute = 180 seconds

song per sample = 24 bits per sample

frequency = 96 kHz

surround sound system = 5.1

solution

first we get here required space for 24 bits at 96 kHz that is

required space = 24 × 96 × 10³ × 6 × 180

required space = 2.48832 × $10^{9}$ bits

as 1 Gib bit = $2^{30}$ bits

so required space = 2.48832 × $10^{9}$ bits ÷ $2^{30}$ bits

required space = 2.3174 Gibibits ...............1

and

space required to save record 16 bit at 44.1 kHz

space required = 16 × 44.1 × 10³ × 3 × 180

space required = 0.381024 × $10^{9}$ bits

space required = 0.381024 × $10^{9}$ bits ÷ $2^{30}$ bits

space required = 0.3548 Gibibits ...........2

so

we get here that save space in 16 bit at 44.1 kHz

save space = 2.3174 Gibibit - 0.3548 Gibibits

save space = 1.9626 Gibibits

8 0

3 years ago