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melomori [17]
3 years ago
12

Suppose you walk into a sauna that has an ambient temperature of 48.0°C. (a) Calculate the rate of heat transfer in watts to you

by radiation given your skin temperature is 37.0°C, the emissivity of skin is 0.98, and the surface area of your body is 1.30 m2.
Physics
1 answer:
GenaCL600 [577]3 years ago
4 0

Answer:99.84 W

Explanation:

Given

Ambient temperature T_{\infty }=48^{\circ}C\approx 321\ K

Temperature of skin T=37^{\circ}C\approx 310\ K

Skin emissivity \epsilon =0.98

Surface area A=1.30\ m^2

Rate of heat transfer is Q=\epsilon A\sigma (T_0^4-T^4)

Q=0.98\times 1.30\times 5.67\times 10^{-8}((321)^4-(310)^4)

Q=99.84\ W

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A vertical steel beam in a building supports a load of 6.0×10⁴. If the length of the beam is 4.0m and it's cross-sectional area
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Answer:

DL = 1.5*10^-4[m]

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With the following equation we can find DL

\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m]

Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.

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A 0.80-kg soccer ball experinces an impulse of 25 N x s . Determine the momentum change of the soccer ball.
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3 0
3 years ago
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At what angle two forces P + Q and (P - Q) act so that their resultant is :
stiv31 [10]

Use resultant formula

\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}

So

#1

A be p+q and B be p-q

\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2

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#2

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)

\\ \rm\Rrightarrow 2cos\beta=0

\\ \rm\Rrightarrow cos\beta=0

\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}

#3

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2

\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)

\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}

\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)

8 0
2 years ago
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