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goldfiish [28.3K]

4 years ago

7

The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The abso

lute pressure at a depth of 5 m in liquid whose specific gravity is 0.8 at the same location.

1 answer:

ser-zykov [4K]4 years ago

8 0

Answer:

a)Patm=135.95Kpa

b)Pabs=175.91Kpa

Explanation:

the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation

Pabs=Pw+Patm(1)

Where

Pabs=absolute pressure

Pw=Water pressure

Patm= atmospheric pressure

Water pressure is calculated with the following equation

Pw=γ.h(2)

where

γ=especific weight of water=9.81KN/M^3

H=depht

A)

Solving using ecuations 1 y 2

Patm=Pabs-Pw

Patm=185-9.81*5=135.95Kpa

B)

Solving using ecuations 1 y 2, and atmospheric pressure

Pabs=0.8x5x9.81+135.95=175.91Kpa

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A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel

Hatshy [7]

Answer:

The load reflection coefficient, $\Gamma =0.62\angle 82.875^{\circ} \Omega$

Reflection coefficient at input, $\Gamma = 0.62\angle - 147.518^{\circ} \Omega$

SWR = 4.26

Given:

Characteristic impedance of the co-axial cable, $Z_{c} = 75 \Omega$

Length of the cable, L = 2.0 cm = 0.02 m

$Z_{Load} = 37.5 + j75 \Omega$

Dielectric constant, K = 2.56

frequency, f = 3.0 GHz = $3.0 \times 10^{9} Hz$

Explanation:

In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance $Z_{i}$ is given by:

$Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}$ (1)

Now, we calculate the value of $\beta$ :

$\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}$

(since, $\lambda' = \farc{c}{f\sqrt{K}}$ )

$\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53$

Now, Substituting the value in eqn (1):

$Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega$

Now, the load reflection coefficient is given by:

$\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}$

Thus

$\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega$

Similarly,

Reflection coefficient at input:

$\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}$

$\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega$

Now, the SWR is given by:

SWR, Standing Wave Ratio = $\frac{1 +|\Gamma|}{1 - |\Gamma|}$

SWR = $\frac{1 +|0.62|}{1 - |0.62|} = 4.26$

8 0

3 years ago

Explain how smart materials can be used by manufacturers to improve health and safety for children's products and goods.

Ierofanga [76]

...simplify devices, reducing weight and the chance of failure.

6 0

2 years ago

A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$

r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Now by putting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

Ф=18.03°

So the shear angle is 18.03°.

4 0

4 years ago

Given the complex numbers A1 5 6/30 and A2 5 4 1 j5, (a) convert A1 to rectangular form; (b) convert A2 to polar and exponential

Deffense [45]

This question is incomplete, the complete question is;

Given the complex numbers A₁ = 6∠30 and A₂ = 4 + j5;

(a) convert A₁ to rectangular form

(b) convert A₂ to polar and exponential form

(c) calculate A₃ = (A₁ + A₂), giving your answer in polar form

(d) calculate A₄ = A₁A₂, giving your answer in rectangular form

(e) calculate A₅ = A₁/( $A^{*}$ ₂), giving your answer in exponential form.

Answer:

a) A₁ in rectangular form is 5.196 + j3

b) value of A₃ in polar form is 12.19∠41.02°

The polar form of A₂ is 6.403 ∠51.34°, exponential form of A₂ = 6.403 $e^{j51.34 }$

c) value of A₃ in polar form is 12.19∠41.02°

d) A₄ in rectangular form is 5.784 + j37.98

e) A₅ in exponential form is 0.937 $e^{j81.34 }$

Explanation:

Given data in the question;

a) A₁ = 6∠30

we convert A₁ to rectangular form

so

A₁ = 6(cos30° + jsin30°)

= 6cos30° + j6cos30°

= (6 × 0.866) + ( j × 6 × 0.5)

A₁ = 5.196 + j3

Therefore, A₁ in rectangular form is 5.196 + j3

b) A₂ = 4 + j5

we convert to polar and exponential form;

first we convert to polar form

A₂ = √((4)² + (5)²) ∠tan⁻¹( $\frac{5}{4}$ )

= √(16 + 25) ∠tan⁻¹( 1.25 )

= √41 ∠ 51.34°

A₂ = 6.403 ∠51.34°

The polar form of A₂ is 6.403 ∠51.34°

next we convert to exponential form;

A∠β can be written as A $e^{j\beta }$

so, A₂ in exponential form will be;

A₂ = 6.403 $e^{j51.34 }$

exponential form of A₂ = 6.403 $e^{j51.34 }$

c) A₃ = (A₁ + A₂)

giving your answer in polar form

so, A₁ = 6∠30 = 5.196 + j3 and A₂ = 4 + j5

we substitute

A₃ = (5.196 + j3) + ( 4 + j5)

= 9.196 + J8

next we convert to polar

A₃ = √((9.196)² + (8)²) ∠tan⁻¹( $\frac{8 }{9.196}$ )

A₃ = √(84.566416 + 64) ∠tan⁻¹( 0.8699)

A₃ = √148.566416 ∠41.02°

A₃ = 12.19∠41.02°

Therefore, value of A₃ in polar form is 12.19∠41.02°

d) A₄ = A₁A₂

giving your answer in rectangular form

we substitute

A₄ = (5.196 + j3) ( 4 + j5)

= 5.196( 4 + j5) + j3( 4 + j5)

= 20.784 + j25.98 + j12 - 15

A₄ = 5.784 + j37.98

Therefore, A₄ in rectangular form is 5.784 + j37.98

e) A₅ = A₁/( $A^{*}$ ₂)

giving your answer in exponential form

we know that $A^{*}$ ₂ is the complex conjugate of A₂

so

$A^{*}$ ₂ = (6.403 ∠51.34° )*

= 6.403 ∠-51.34°

we convert to exponential form

A∠β can be written as A $e^{j\beta }$

$A^{*}$ ₂ = 6.403 $e^{-j51.34 }$

also

A₁ = 6∠30

we convert to polar form

A₁ = 6 $e^{j30 }$

so A₅ = A₁/( $A^{*}$ ₂)

A₅ = 6 $e^{j30 }$ / 6.403 $e^{-j51.34 }$

A₅ = (6/6.403) $e^{j(30+51.34) }$

A₅ = 0.937 $e^{j81.34 }$

Therefore A₅ in exponential form is 0.937 $e^{j81.34 }$

6 0

3 years ago

you are feeling lazy and you binge watch netflix for 2 hours lying in bed. your laptop's power draw is 60w from it's fully charg

Leni [432]

The Energy discharged from the battery within 2 hours is 0.12kWh

<h3>Energy</h3>

Energy is the capacity for doing work. It may exist in p<em>otential, kinetic, thermal, electrical, chemical, nuclear</em>. It is measured in joules.

A kilowatt hour (kWh) is a measure of how much energy you're using per hour.

From the question:

Energy discharged from the battery = 60W * 2 hours = 120 Wh = 0.12kWh

The Energy discharged from the battery within 2 hours is 0.12kWh

Find out more on energy at: brainly.com/question/582060

8 0

3 years ago

Read 2 more answers