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brilliants [131]

2 years ago

8

If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus

e your skin is usually a little colder than your insides), what is the total power that person will radiate? For now, assume the person is a perfect blackbody (so ϵ=1).

1 answer:

anastassius [24]2 years ago

5 0

Answer:

$E=477.92\ W.m^{-2}$

Explanation:

Given that:

Absolute temperature of the body, $T=273+30=303\ K$

emissivity of the body, $\epsilon=1$

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

$E=\epsilon. \sigma.T^4$

where:

$\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}$ (Stefan Boltzmann constant)

Now putting the respective values:

$E=1\times 5.67\times 10^{-8}\times 303^4$

$E=477.92\ W.m^{-2}$

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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time $t_{arrow}$ travels the distance d with a speed $v_{arrow}$ of 40 m/s and hits the target. We can see that the equation will be:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at 340 m/s in a time $t_{sound}$ . Obtaining :

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

$t_{arrow} + t _{sound} = 1 s$ .

This equation allows us to write:

$t _{sound} = 1 s - t_{arrow}$ .

Plugging this relationship in the distance equation for the sound:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Now, we can replace d from the first equation, and obtain:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, we can just work a little bit:

$40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s$ .

Now, we can just plug this value into the first equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

6 0

2 years ago

An airplane weighing 5000 lb is flying at standard sea level with a velocity of 200 mi/h. At this velocity the L/D ratio is a ma

saul85 [17]

Answer:

98.15 lb

Explanation:

weight of plane (W) = 5,000 lb

velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s

wing area (A) = 200 ft^{2}

aspect ratio (AR) = 8.5

Oswald efficiency factor (E) = 0.93

density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}

Drag = 0.5 x ρ x $v^{2}$ x A x Cd

we need to get the drag coefficient (Cd) before we can solve for the drag

Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)

where

induced drag coefficient (Cdi) = $\frac{Cl^{2} }{n.E.AR}$ (take note that π is shown as n and ρ is shown as $p$ )

where lift coefficient (Cl)= $\frac{2W}{pAv^{2} }$ = $\frac{2x5000}{0.002377x200x293.3^{2} }$ = 0.245

therefore

induced drag coefficient (Cdi) = $\frac{Cl^{2} }{n.E.AR}$ = $\frac{0.245^{2} }{3.14x0.93x8.5}$ = 0.0024

since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)

Cd = 0.0024 + 0.0024 = 0.0048

Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.

Drag = 0.5 x ρ x $v^{2}$ x A x Cd

Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb

8 0

2 years ago

A friend of yours who takes her astronomy class very seriously challenges you to a contest to find the thinnest crescent moon yo

RUDIKE [14]

Answer:

after the sun sets or just as it is setting

Explanation:

a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.

8 0

2 years ago

A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release

Scilla [17]

Answer:

(A) As it moves farther and farther from Q, its speed will keep increasing.

Explanation:

When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.

Mathematically:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}$

where:

r = distance between the charges

$\epsilon_0=$ permittivity of free space

By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.

7 0

2 years ago

In which of the following examples does the object have both kinetic and potential energy? Select all that apply.

notsponge [240]

I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic

6 0

2 years ago

Read 2 more answers