Two infinite extent current sheets exist at z = −3.0 m and at z = +3.0 m. The top sheet has a uniform current

) A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging proc

bonufazy [111]

Answer:

it will be a scattered plot

Explanation:

5 0

2 years ago

2. A F-22 Raptor has just climbed through an altitude of 9,874 m at 1,567 kph when a disk

BabaBlast [244]

The pressure difference across the sensor housing will be "95 kPa".

According to the question, the values are:

Altitude,

9874

Speed,

1567 kph

Pressure,

122 kPa

The temperature will be:

→ $T = 15.04-[0.00649(9874)]$

→ $= 15.04-64.082$

→ $= -49.042^{\circ} C$

now,

→ $P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}$

→ $= 27.074$

hence,

→ The pressure differential will be:

= $122-27$

= $95 \ kPa$

Thus the above solution is correct.

Learn more about pressure difference here:

brainly.com/question/15732832

3 0

2 years ago

What did August Comte contribute to sociology including positivism

Travka [436]

Answer:

Auguste Comte was the first to develop the concept of "sociology." He defined sociology as a positive science. Positivism is the search for "invariant laws of the natural and social world." Comte identified three basic methods for discovering these invariant laws, observation, experimentation, and comparison.

Explanation:

I hope it's help u :)

4 0

2 years ago

Risks in driving never begins with yourself, but with other drivers who take risks.

Ymorist [56]

False! Just saying. You could be under the influence, or just have no clue as to what you're doing.

8 0

2 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago