Which of the following are located on a plant's leaves?

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0

3 years ago

Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.

Sati [7]

Answer:

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$ , $z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$ , $z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

$z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]$

Where angles are measured in radians and k represents an integer between $0$ and $n - 1$ .

The magnitude of the complex number is $27$ and the equivalent angular value is $1.817\pi$ . The set of cubic roots are, respectively:

k = 0

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$

k = 1

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$

k = 2

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

5 0

3 years ago

The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas

alexgriva [62]

Answer:

B. inverse plot, 0.51 kilograms/meter3

Explanation:

First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

$y \propto \frac{1}{x}$

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B. inverse plot, 0.51 kilograms/meter3

5 0

3 years ago

REALICE UN ANALISIS DEL TEMA, INDICANDO LAS IDEAS PRINCIPALES

katen-ka-za [31]

Answer:

De cual Tema??? jajaja no se que necesitas

8 0

3 years ago

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Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0

3 years ago

Read 2 more answers