Solve the compound inequality. 3x − 4

18

a_sh-v [17]

Answer:

englishhhhhhhhh

Explanation:

we dont understand

6 0

3 years ago

The thrust F of a screw propeller is known to depend upon the diameter d,speed of advance \nu ,fluid density p, revolution per s

musickatia [10]

Answer:

screw thrust = ML $T^{-2}$

Explanation:

thrust of a screw propeller is given by the equation = p $V^{2}$ $D^{2}$ x $\frac{ND}{V}$ Re

where,

D is diameter

V is the fluid velocity

p is the fluid density

N is the angular speed of the screw in revolution per second

Re is the Reynolds number which is equal to puD/μ

where p is the fluid density

u is the fluid velocity, and

μ is the fluid viscosity = kg/m.s = M $L^{-1}$ $T^{-1}$

Reynolds number is dimensionless so it cancels out

The dimensions of the variables are shown below in MLT

diameter is m = L

speed is in m/s = L $T^{-1}$

fluid density is in kg/ $m^{3}$ = M $L^{-3}$

N is in rad/s = L $L^{-1}$ $T^{-1}$ =

If we substitute these dimensions in their respective places in the equation, we get

thrust = M $L^{-3}$ $(LT^{-1}) ^{2}$ $L^{2}$ $\frac{T^{-1} L}{LT^{-1} }$

= M $L^{-3}$ $L^{2}$ $T^{-2}$

screw thrust = ML $T^{-2}$

This is the dimension for a force which indicates that thrust is a type of force

6 0

3 years ago

. A normal-weight concrete has an average compressive strength of 20 MPa. What is the estimated flexure strength

bulgar [2K]

Answer:

2.77mpa

Explanation:

compressive strength = 20 MPa. We are to find the estimated flexure strength

We calculate the estimated flexural strength R as

R = 0.62√fc

Where fc is the compressive strength and it is in Mpa

When we substitute 20 for gc

Flexure strength is

0.62x√20

= 0.62x4.472

= 2.77Mpa

The estimated flexure strength is therefore 2.77Mpa

4 0

3 years ago

Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at

AysviL [449]

Answer:

(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

State1 : p1 = 100kpa, T1= 25+273 =298k

From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

State 2a:p2=2mpa,t2=540k (actual condition 2a)

h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

=246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

=s20-s10-Rln(p2/p1), T0=250C

=(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

= 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

= -38.39 kJ/kg

a)Reversible work = Win actual -any irreversiblities present

=246.17 + irreversibilty

=246.17+-38.39

=207 kJ/kg

8 0

3 years ago

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these d

Tems11 [23]

Complete Question

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devises are available for weekend warriors who want to play with their chain saws. Let us model the illustrated tongs as a planar mechanism that carries a log of weight 210 N. Given the following dimensions: 35 mm 10 mm 40 mm 230 mm 85 mm 45 mm 10 mm 35 mm 345 mm determine the force in N and moment in Nm that our worker exerts on the tongs. Also determine the pinching force magnitude in N that the tongs exert on the log; i.e. determine the horizontal force that the tong's teeth exert on the log. Assume the point E is centered between the tong's teeth.

The diagram for this question is shown on the first uploaded image

Answer:

The force P is $P= 210 N$