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3 years ago

13

If a machine uses LESS effort to overcome a given resistance force (if Fe is less than FR), it has an actual mechanical advantag

e of
Your answer:
0
Exactly 1
Less than 1
Greater than 1

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

Answer:

greater than 1

Explanation:

The mechanical advantage is the ratio Fr/Fe. So if Fr > Fe, the mechanical advantage is greater than 1.

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For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are

kotegsom [21]

Answer:

B. The thickness of the heated region near the plate is increasing.

Explanation:

First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.

From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.

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3 years ago

La extensión de un resumen debe estar en un rango de _________ del texto inicial. a)25 a 35 % b)10 a 20 % c)15 a 20 % d)20 a 25

mestny [16]

Answer:

b)

Explanation:

because it is correct

3 0

3 years ago

*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t

riadik2000 [5.3K]

Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

B) ε_y = 0.00375

C) γ_xy = 0.0122 rad

Explanation:

We are given;

δ = 0.03 in

L = 4 in

ν_r = 0.5

θ = 89.3° = 89.3π/180 rad

Let's calculate ε_x in the direction of axis x

Thus, ε_x = δ/L = 0.03/4 = 0.0075

Let's calculate ε_y in the direction of axis y;

ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375

Now, shear strain is angle between π/2 rad surfaces at that point.

Thus,

γ_xy = π/2 - θ = π/2 - 89.3π/180

γ_xy = π(0.003889) = 0.0122 rad

3 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

3 years ago

What are two ways you can see that there is a smaller load going to the base of the transistor?

Lyrx [107]

Answer:

<h3><em>Transistor switches can be used to switch a low voltage DC device (e.g. LED’s) ON or OFF by using a transistor in its saturated or cut-off </em><em>state</em></h3>

<em>. Cut-off </em><em>Region</em>

<em>Here the operating conditions of the transistor are zero input base current ( IB ), zero output collector current ( IC ) and maximum collector voltage ( VCE ) which results in a large depletion layer and no current flowing through the device. Therefore the transistor is switched “Fully-OFF”.</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>2</em><em>.</em><em>Saturation </em><em>Region</em>

<em>Here the transistor will be biased so that the maximum amount of base current is applied, resulting in maximum collector current resulting in the minimum collector emitter voltage drop which results in the depletion layer being as small as possible and maximum current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.</em>

Explanation:

hope it helps

7 0

2 years ago