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3 years ago

10

If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of

the pulse if you stretch the hose more tightly

1 answer:

stich3 [128]3 years ago

5 0

Answer:

Explanation:if you stretch the hose more tightly the speed of the pulse will reduce..

You might be interested in

what do we call a landslide in which fine-grained soil moves cohesively but with extensive internal shearing?

pogonyaev

It is called an earth flow

5 0

3 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

3 years ago

A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP

Aleks04 [339]

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

<u>a) Determine the pressure at inlet to expansion process</u>

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp) $\frac{r-1}{r}$ ]

0.569 = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569 = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

= 5.2 kPa

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

= 0.493 = 49.3%

4 0

3 years ago

Shortly after the introduction of a new coin, newspapers published articles claiming the coin is biased. The stories were based

garik1379 [7]

Answer:

(a) 0.12924

(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

Explanation:

(a)

n=200 for fair coin getting head, p= 0.5

Expectation = np =200*0.5=100

Variance = np(1 - p) = 100(1-0.5)=100*0.5=50

Standard deviation, $s = \sqrt {variance}=\sqrt {50}= 7.071068$

Z value for 108, $z =\frac {108-100}{7.071068}= 1.131371$

P( x ≥108) = P( z >1.13)= 0.12924

(b)

Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

3 0

3 years ago

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough

jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection

Explanation:

We are given:

Plane strain fracture toughness K $= 98.9 MPa \sqrt{m}$

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

$oc = \frac{K}{Y \sqrt{pi * a}}$

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

$a= \frac{1}{pi} * [\frac{k}{y*a}]^2$

Substituting figures in the expression above, we have:

$= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2$

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection

3 0

3 years ago

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