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2 years ago

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Physics

1 answer:

BARSIC [14]2 years ago

5 0

Answer:

The correct answer is Tor buni aga daka pagli Tor buda mu tor buni kaitam chai ami

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

juin [17]

Answer:

The maximum temperature is 90.06° C

Explanation:

Given that

t= 0.1 mm

Heat generation

$q_g=0.3\ MW/m^3$

Heat transfer coefficient

$h=500\ W/m^2K$

Here one side(left side) of the wall is insulated so the all heat will goes in to right side .

The maximum temperature will at the left side.

Lets take maximum temperature is T

Total heat flux ,q

$q=q_g\times t$

$q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2$

$q=30\ W/m^2$

So the total thermal resistance per unit area

$R=\dfrac{t}{K}+\dfrac{1}{h}$

$R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}$

R=0.002 K/W

We know that

q=ΔT/R

30=(T-90)/0.002

T=90.06° C

The maximum temperature is 90.06° C

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2 years ago

At the same instant that a 0.50-kg ball is dropped from 25m above Earth,? At the same instant that a 0.50-kg ball is dropped fro

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I hope it is clearly visible.. Velocity of the center of mass of 2-ball system is - 11.54m/s. Minus indicates, velocity direction is in downward direction.

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How people get along with one another is called

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2 years ago

Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

$\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}$

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

$\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}$

$\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}$

$d=35.35\ m$

The range is 35.35 m

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2 years ago