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azamat
3 years ago
9

The Fisher effect says that _______ . Group of answer choices the nominal interest rate adjusts one for one with the inflation r

ate the growth rate of the money supply is negatively related to the velocity of money real variables are heavily influenced by the monetary system All of the above are correct.
Engineering
1 answer:
Greeley [361]3 years ago
7 0

Answer:

what wrong subject

Explanation:

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Can you guys please introduce yourself​
EleoNora [17]

Answer: why?

Explanation:

6 0
3 years ago
Read 2 more answers
If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
Stolb23 [73]

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

3 0
3 years ago
A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete
Alex

Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

d is grain diameter to be achieved after heating in mm

d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

t₁= 500 min = 500/60 =25/3 hrs

t₂=?

n=2.2

First find K

K=(d²-d²₀ )/t₁

K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3

K=(0.051²-0.024²) ÷25/2

K=0.000243 mm²/h

Re-arrange equation for K ,to get the equation for d as;

d=√(d₀²+ Kt)  where now t=t₂

d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min

4 0
3 years ago
A segment of four-lane freeway (two lanes in each direction) has a 3% upgrade that is 1500 ft long followed by a 1000-ft 4% upgr
Dennis_Churaev [7]

Answer:

The level of service of of compound grade freeway is LOSB.

Explanation:

Find the provided attachments for explanation

3 0
3 years ago
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
4 years ago
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