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Elanso [62]
3 years ago
8

Three spheres are subjected to a hydraulic stress. The pressure on spheres 1 and 2 is the same, and they are made of the same ma

terial. The fractional compression on the third sphere is equal to the fractional compression on the first sphere times the reciprocal of the fractional compression on the second. If the pressure on the third sphere is 150000 N/m2 , what is the Bulk Modulus for the third sphere
Engineering
1 answer:
r-ruslan [8.4K]3 years ago
5 0

Answer:

"150000 N/m²" is the right approach.

Explanation:

According to the question, the pressure on the two spheres 1 and 2 is same.

Sphere 1 and 2:

Then,

⇒  P_1=P_2

⇒  \frac{\Delta V_1}{V_1}=\frac{\Delta V_2}{V_2}

and the bulk modulus be,

⇒  B_1=B_2

Sphere 3:

⇒  \frac{\Delta V_3}{V_3} =\frac{\frac{\Delta V_1}{V_1} }{\frac{\Delta V_2}{V_2} } =1

then,

⇒  P_3=B\times \frac{\Delta V_3}{V_3}

⇒       =B\times 1

⇒       =150000\times 1

⇒       =150000 \ N/m^2

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At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge
ale4655 [162]

Answer:

<em>a) 50 J/kg</em>

<em>b) 721 67 KW</em>

<em></em>

Explanation:

The velocity of the wind v = 10 m/s

diameter of the blades d = 70 m

efficiency of the turbine η = 30%

density of air ρ = 1.25 kg/m^3

The area of the blade A = \pi d^2/4

A = \frac{3.142 * 70^2}{4} = 3848.95 m^2

The mechanical energy air per unit mass is gives as

e = v^2/2 = \frac{10^2}{2} = <em>50 J/kg</em>

<em></em>

Theoretical Power of the turbine P = ρAve

where

ρ is the density of air

A is the area of the blade

v is the velocity of the wind

e is the energy per unit mass

substituting values, we have

P = 1.25 x 3848.95 x 10 x 50 = 2405593.75 W

Actual power = ηP

where η is the efficiency of the turbine

P is the theoretical power of the turbine

Actual power = 0.3 x 2405593.75 = 721678.1 W

==> <em>721 67 KW</em>

7 0
3 years ago
In multi-grade oil what is W means?
irga5000 [103]

Answer:

winter viscosity grades

Explanation:

The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.

4 0
3 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
3 years ago
Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.
mezya [45]
I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!
4 0
3 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0
3 years ago
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