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3 years ago

Determine the mass of a car that weight 3,500 lbs both in slugs and kilograms. The answer should be in kilograms.

Engineering

1 answer:

Andreas93 [3]3 years ago

4 0

Answer:

The mass of car in slug =108.5 slug.

The mass of car in kilogram =1575 Kg.

Explanation:

Given that

Mass of car = 3500 lbs

We know that

1 lbs =0.031 slug

3500 lbs = 0.031 x 3500 slug

So the mass of car in slug =108.5 slug.

We know that

1 lbs =0.45 kilograms

3500 lbs = 3500 x 0.45 Kg

So the mass of car in kilogram =1575 Kg.

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Technicians have to determine the flow rate of water in a pipe with the aid of a venturi installation and a mercury au.oaeter. T

Schach [20]

Answer:

Manometric difference x=142.85 mm.

Explanation:

Given :

Pipe diameter $d_1=40 mm$

venturi meter $d_2=20 mm$

We can know that discharge through venturi meter is given as

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt {A_1^2-A_2^2}}$

$A_1=1.24\times 10^{-3},A_2=3.12\times 10^{-4}$

$Q=A_1V_1$

$Q=1.24\times 10^{-3}\times 1.5=0.00186 m^3/s$

$0.00186=0.97\dfrac{1.24\times 10^{-3}\times 3.12\times 10^{-4} \sqrt{2gh}}{\sqrt {(1.24\times 10^{-3})^2-(3.12\times 10^{-4})^2}}$

h=1.8 m

We know that $h=x\left (\dfrac{\rho_{hg}}{\rho_w}-1\right )$

Where x is the manometric deflection

⇒ $1.8=x\left (\dfrac{13600}{1000}-1\right )$

So x=14.28 mm

Manometric difference x=142.85 mm.

7 0

3 years ago

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

8 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

enyata [817]

Answer:

Check image, right on PLATO

Explanation:

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3 years ago

Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Dr

Troyanec [42]

Complete Question:

Show the bias polarities and depletion regions of an npn BJT in the normal active, saturation, and cutoff modes of operation. Draw the three sketches one below the other to (qualitatively) reflect the depletion widths for these biases, and the relative emitter, base, and collector doping.

Consider a BJT with a base transport factor of 1.0 and an emitter injection efficiency of 0.5.

Calculate roughly by what factor would doubling the base width of a BJT would increase, decrease, or leave unchanged the emitter injection efficiency and base transport factor? Repeat for the case of emitter doping increased 5 × =. Explain with key equations, and assume other BJT parameters remain unchanged!

Answer & Explanation:

[Find the attachments]

Step 1 :

Emitter and base, collector, and base are forward biased then BJT is in saturation region. Emitter and base is forward biased and base and collector in reverse biased then BJT is in active region.

Emitter and base, collector and base are reverse biased then BJT in cut off region.

Three sketches one below the other is shown in Figure 1.

[find the figure in attachment]

Step 2:

Value of base widths of saturation, active and cut off operated BJT are value of Base width of saturated region operated BJT is less than base width in active region operated BJT. Value of base width of active region operated BJT is less than base width in cut off region operated BJT.

Saturation region operated base width of BJT is < Active region operated base width of BJT is < Cut off region operated base width of BJT.

[For Steps 3 4 5 6 and 7 find attachments]

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4 years ago

When using fall arrest, free fall must be kept at or below how many feet

SashulF [63]

<h3>Answer:</h3>

two feet or less

<h3>Explanation:</h3>

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3 years ago