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3 years ago

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which one of the following appliance parts gets the hardest services? A. Heating elementwhich one of the following appliance par

ts gets the hardest services? A. Heating element

2 answers:

makvit [3.9K]3 years ago

8 0

The following appliance parts gets the hardest services is a line cord.
The answer is D.
(U didn’t write the right.)

padilas [110]3 years ago

8 0

The answer is D, hope this helps, good luck

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EMB agar is a medium used in the identification and isolation of pathogenic bacteria. It contains digested meat proteins as a so

Ilya [14]

Answer:

A selective medium, a differential medium, and a complex medium.

Explanation:

A selective media is a microbiological media which only support the growth of a particular specie or types of species of microorganisms,this media acts in such a way to inhibit or hinder the growth of other microorganisms.

Differential media are media that acts to Identifying particular strains of microorganisms of similar species.

Complex media are media used for the growth of microorganisms this which contains complex or a wide range of nutrients with chemical composition which may be difficult to determine.

5 0

3 years ago

A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf

galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D ) = 0.25 m

Temperature of sphere( T ) = 35° C

Temperature of surrounding ( T∞ ) = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = $\pi D^{2}$

= $\pi * 0.25^2$ = 0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( $T^{4} - T^{4} _{s}$ ) ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

= 22.83 watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

= 11*0.2 ( 35 -25 ) = 22 watts

Hence total rate of heat loss

= 22 + 22.83

= 44.83 watts

5 0

3 years ago

Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held

77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }$

${T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K$

The heat supplied = $\dot {m}_f$ × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = $\dot m$ · $c_p(T_3 - T_2)$

$\dot m$ = 20 kg/s

The heat supplied = 20* $c_p(T_3 - T_2)$ = 21,350 kJ/s

$c_p$ = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

$\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}$

T₅ = 1208.42*(2/2.333) = 1035.94 K

$C_j = \sqrt{\gamma \times R \times T_5}$ = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = $\dot m$ × $C_j$ = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0

3 years ago

A two-dimensional flow field described by

Oduvanchick [21]

Answer:

the answer is

Explanation:

<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>

5 0

3 years ago

Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r

My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0

3 years ago