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2 years ago

HELP ASAP PLZ!

Physics

2 answers:

Dafna1 [17]2 years ago

8 0

1. is D. Scientific method cannot be replaced.
2. is B.
3. is B as well.

Sveta_85 [38]2 years ago

6 0

For #1 the answer that is not true is as a computer simulation to predict weather then for #2 the answer is The process that scientists use to gain knowledge then for #3 the answer is to test a hypothesis

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Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

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3 years ago

Add the vectors:

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Vector 1 has components

$x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m$

$y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m$

and vector 2 has

$x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m$

$y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m$

Add these vectors to get the resultant, which has components

$x_{\rm total}\approx11.133\,\mathrm m$

$y_{\rm total}\approx13.268\,\mathrm m$

The magnitude of the resultant is

$\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m$

with direction $\theta$ such that

$\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ$

or about 50º N of E.

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3 years ago