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3 years ago

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What is the net power that a person with surface area of 1.20 m2 radiates if his emissivity is 0.895, his skin temperature is 27

°C, and he is in a room that is at a temperature of 17°C m2 radiates if his emissivity is 0.895, his skin temperature is 27°C, and he is in a room that is at a temperature of 17°C. (σ = 5.67 × 10-8 W/m2 ∙ K4)

1 answer:

marissa [1.9K]3 years ago

4 0

Explanation:

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A ship leaves a port at noon and travels due west at 20 knots. At 6 PM, a second ship leaves the same port and travels northwest

Amanda [17]

Answer:

$v = 12.44 Knots$

Explanation:

First ship starts at Noon with speed 20 Knots towards West

now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West

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$r_1 = 20(t + 6)\hat i$

$r_2 = 15(t)(cos45\hat i + sin45\hat j)$

now we can find the distance between two ships as

$x = \sqrt{(20(t + 6) - 10.6 t)^2 + (10.6t)^2}$

now we have

$x^2 = (120 + 9.4 t)^2 + (10.6 t)^2$

$x^2 = 200.72 t^2 + 14400 + 2256 t$

now we will differentiate it with respect to time

$2x\frac{dx}{dt} = 401.44 t + 2256$

here we know that

$t = \frac{90}{15} = 6 hours$

so we have

$x = 187.5$

now we have

$2(187.5) v = 401.44(6) + 2256$

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3 years ago

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Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

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$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

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$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

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It is called SUBLIMATION.

so the correct answer is (D).

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