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Nataliya [291]
3 years ago
9

Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation, in kW, from 1

.02 m2 of 1000ºC fresh lava into 25.3ºC surroundings, assuming lava’semissivity is 0.97.
Physics
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

1.5 x 10⁵ W

Explanation:

A = Area of the fresh lava = 1.02 m²

T = Temperature of fresh lava = 1000 °C = 1000 + 273 = 1273 K

T₀ = Temperature of surrounding = 25.3 °C = 25.3 + 273 = 298.3 K

ε = emissivity of the lava = 0.97

σ = stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Rate of  transfer is given as

E = σ ε A (T⁴ - T₀⁴)

E = (5.67 x 10⁻⁸) (0.97) (1.02) ((1273)⁴ - (298.3)⁴)

E = 1.5 x 10⁵ W

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Which of the following is true about this lever?
zysi [14]

As we can see here the lever has two forces at two ends

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2. counterclockwise torque due to force at left end

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Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was
tigry1 [53]

Answer:

\Delta V=316167V

Explanation:

The difference of electric potential between two points is given by the formula \Delta V=Ed, where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is E=\frac{F}{q}, where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have \Delta V=\frac{Fd}{q}, so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force N=W-F. On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) d=\frac{at^2}{2}, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

a_1=\frac{2d}{t_1^2}

a_2=\frac{2d}{t_2^2}

These relate with the forces by Newton's 2nd Law:

W=ma_1

N=ma_2

Putting all together:

N=W-F=ma_1-F=ma_2

Which means:

F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})

And finally we substitute:

\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})

Which for our values means:

\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V

7 0
3 years ago
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