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2 years ago

A 2.8 kg rectangular air mattress is 2.00 m long, 0.500 m wide, and

1 answer:

8 0

First off, we have to assume that the 2.8 kg mass of the air matress includes both the material that the matress is constructed from and the air that is contained within.

The mattress can support a total weight that is equal to the weight of water that the air matress displaces.

The volume of the air mattress is:

V(mattress) = (2 m)(0.5 m)(0.1 m) = 0.1 m³

This volume is equal to the maximum amount of water that the matress can displace in the water before it begins sinking.

The density of water is 1000 kg / m²

The weight of the maximum amount of displaced water is:

V(matress) * (density of water) * (g) = (0.1 m³) * (1000 kg / m³) * (9.81 m / s²) = 981 N.

Let m represent the mass that the matress is supporting.

The total weight of the matress and the supported mass is:

(2.8 + m) * 9.81

Setting this weight equal to the weight of the water displaced gives:

(2.8 + m) * 9.81 = 981

2.8 + m = 100

m = 100 - 2.8

m = 97.2 kg

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Explain why a roller coaster’s second hill cannot be taller than the first hill.

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There will not be enough momentum from the first hill to cross another hill if he same or larger size because of the way potential energy and kinetic energy works it will not be able go as high as it could go on he fist hill.

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2 years ago

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Green plants convert light energy from the sun into ______. * a. gravitational potential energy b. chemical potential energy c.

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Answer:

chemical potiential energy

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2 years ago

26. Keenan found the mass of a book to be 4.56*10^ -2 kg . What is the mass of the book in milligrams?

vagabundo [1.1K]

Taking into account the rule of three for the change of units, the mass of the book is 45600 miligrams.

First of all, the rule of three is a mathematical tool that helps you quickly solve proportionality problems.

Having three known values and one unknown, a proportional relationship is established between all of them in order to find the fourth term of the proportion.

If the relationship between the magnitudes is direct (when one magnitude increases, so does the other; or when one magnitude decreases, so does the other), the rule of three is applied as follows, where a, b and c are known values and x is the unknown to calculate:

a → b

c → x

So: $x=\frac{cxb}{a}$

Being 1 kg equivalent to 1000000 milligrams, In this case the rule of three is applied as follows: if 1 kg equals 1000000 milligrams, 4.56×10⁻² kg equals how many milligrams?

1 kg → 1000000 milligrams

4.56×10⁻² kg → x

So:

$x=\frac{4.56x10^{-2} kg x1000000 miligrams }{1 kg}$

<u><em>x=45600 miligrams</em></u>

In summary, the mass of the book is 45600 miligrams.

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2 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

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2 years ago

Which statement is correct?

miv72 [106K]

'A' is correct. B, C, and D are false statements.

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2 years ago

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