What is a relatively soft layer of upper mantle called?

The heat flux that is applied to one face of a plane wall is q″ = 20 W/m2. The opposite face is exposed to air at temperature 30

Blababa [14]

Answer:

400 W/m^2 and 31℃

Explanation:

The output heat flux q"= 20 W/m^2 (geven)

The output heat flux from.the wall to the air by convection

q"conv = h(ts - t∞)

q"conv = 20(50-30) = 400 W/m^2

Therefor, this case is unsteady and the wall temperature changes with time till the energy balance exist.

ENERGY BALANCE

The input energy must be equal to the output energy for steady state condition. If not the state will be unstaidy or transient.

2. Its noticed that the output heat flux is not that the I put heat flux, therefore the wall tempers will be decreased till the output heat flux is reduced to the value of the given input heat flux

T steady = T∞ +q"/h

= 30 + 20/20 = 31℃

3 0

3 years ago

PLEASE HELP 15 POINTS If you hold a bar magnet in each hand and bring your hands together, will the force be attractive or repul

marusya05 [52]

Answer:

with the 2 north poles it will repulse with the north and south it will retract

Explanation:

north and north wont come together. same for south and south. opposite sides always attract one another. hope it helped!!!

6 0

2 years ago

You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The

Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is $h=3.6\ m$

Initial speed of Putty $u=9.5\ m/s$

Speed of Putty just before it strike the ceiling is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-9.5^2=2\times (-9.8)\times 3.6$

$v^2=19.69$

$v=4.43\ m/s$

time taken by putty to reach the ceiling

$v=u+at$

$4.43=9.5-9.8\times t$

$t=\frac{5.07}{9.8}$

$t=0.517\ s$

8 0

3 years ago

12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car c

Triss [41]

Answer:

22.5 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Time (t) = 1.5 s

Final velocity (v) = 0 m/s

Distance (s) =?

The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:

s = (u + v)t/2

s = (30 + 0)1.5 / 2

s = (30 × 1.5) / 2

s = 45 / 2

s = 22.5 m

Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.

4 0

3 years ago

A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0

2 years ago