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3 years ago

Fluids at rest possess no flow energy. a)- True b)- False

Engineering

1 answer:

gtnhenbr [62]3 years ago

8 0

Answer:

True.

Explanation:

According the engineering flow they don not possess flow energy when they are in rest.

When they are in motion they show a translation energy.

The features if fluids may be different according the variables of pressure and temperature.

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A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus

iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

$\sigma = \frac{F}{A}\\\\$

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5 $\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2$

Therefore,

$\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips$

Now, we determine the strain:

$strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}$

Now, the modulus of elasticity (E) is given as:

$E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}$

7 0

3 years ago

How a single force is resolved along the perpendicular axis acting an angle theta with horizontal?

Nataly [62]

Answer:

A single force, which is acting at angle θ from a horizontal axis, can be resolved into components which act along the perpendicular axis.

Consider the perpendicular axis x and y, where x represents the horizontal axis and y represents vertical axis.

The Force is resolved into 2 parts, one acts along x-axis and is represent by X. The other acts along y-axis and is represented by Y.

From the diagram we can see that the Force and its components X and Y makes up a right angles triangle, where θ is the angle from the x-axis

We know that:

cosθ = Base/Hypotenuse

cosθ = X/F

X = Fcosθ

We know that:

sinθ = Perpendicular/Hypotenuse

sinθ = Y/F

Y = Fsinθ

<h3>Relation of Force and its Components:</h3>

Force F can be represent by:

F = Fcosθ (along x-axis) + Fsinθ (along y-axis)

As they form a right angled triangle, we can use Pythagoras Theorem to show the relation between Force and its components.

Hypotenuse² = Base² + Perpendicular²

F² = X² + Y²

F² = (Fcosθ)² + (Fsinθ)²

$F = \sqrt{(F\cos\theta)^2+(Fsin\theta)^2}$

Where θ can be found by using any of the trignometric functions.

6 0

3 years ago

What is the length of x

kkurt [141]

Answer:

14/64

Explanation:

8 0

3 years ago

A series resistive circuit has two resistors. R1 is 570 ohms and R2 is 560 ohms.

ZanzabumX [31]

Answer:

10.203 Volts

Explanation:

For this problem, we need to understand that a series resistive circuit is simply a circuit with some type of voltage source and some resistors, in this case, R1 and R2.

First, we need to find the voltage in the circuit. To do this, we need to find the total resistance of the circuit. When two resistors are in series, you sum the resistance. So we can say the following:

R_Total = R1 + R2

R_Total = 570 Ω + 560 Ω

R_Total = 1130 Ω

Now that we have R_Total for the circuit, we can find the voltage of the circuit by using Ohm's law, V = IR.

V_Total = I_Total * R_Total

V_Total = 17.9 mA * 1130 Ω

V_Total = 20.227 V

Now that we have V_Total, we can find the voltage drop across each resistor by using Ohm's law once more. Note, that since our circuit is series, both resistors will have the same current (I.e., I_Total = I_1 = I_2).

V_Total = V_1 + V_2

V_Total = V_1 + I_2*R2

V_Total - I_2*R2 = V_1

20.227 V - (17.9 mA * 560 Ω) = V_1

20.227 V - (10.024 V) = V_1

10.203 V = V_1

Hence, the voltage drop across R1 is 10.203 Volts.

Cheers.

3 0

3 years ago

Liquid water enters a valve at 300 kPa and exits at 275 kPa. As water flows through the valve, the change in its temperature, st

Elan Coil [88]

Answer:

The change in kinetic energy per unit mass of water flowing through the valve is - ΔKE = 0.025 KJ/Kg

Explanation:

Knowing

-Fluid is air

-inlet 1: P1 = 300 kPa

-exit 2: P2 = 275 kPa

density - rho= 1000 kg/m3

Using the formula

Δh = cΔT + Δp/rho

as change in temperature is neglected then change in enthalpy becomes

Δh = Δp/rho

energy equation could be defined by

Q - W = m(out) [h(out) $V^{2}$ (out)/2 + g Z(out)] - m(in) [h(in) $V^{2}$ (in)/2 + g Z(in)]

Q - W = m2 [h2 $V^{2}$ 2/2 + g Z2] - m1 [h1 $V^{2}$ 1/2 + g Z1]

as for neglecting potential energy effects

Q - W = m2(h2) - m1(h1)

as the system is adiabatic and has no work done

0 = m2 [h2 $V^{2}$ 2/2] - m1 [h1 $V^{2}$ 1/2]

from mass balance m1 = m2

0 = [h2 $V^{2}$ 2/2] - [h1 $V^{2}$ 1/2]

Change in kinetic energy could be defined by

ΔKE = $V^{2}$ 2/2 - $V^{2}$ 1/2

Change in specific enthalpy could be defined by

Δh = h2 - h1

Then the change in kinetic energy per unit mass of water flowing through the valve could be calculated as following

ΔKE = -Δh = ΔP/rho

-(275 - 300)/1000 = 0.025 KJ/Kg

- ΔKE = 0.025 KJ/Kg

3 0

4 years ago