Why is the tropical climate warmer than the climate at the north and south poles?

A 1.0 m × 1.5 m double-pane window consists of two 20-mm-thick layers of glass (k = 0.78 W/m·K) that are separated by a 13-mm ai

jeka94

Answer:

$\dot Q=105.042\ W$ is the heat loss by conduction.

Explanation:

Given:

area of window pane, $A=1.5\ m^2$

thickness of the glass pane, $x=0.02\ mm$

thickness of the air gap, $x'=0.013\ m$

thermal conductivity of the glass, $k=0.78\ W.m^{-1}.K^{-1}$

thermal conductivity of the air, $k_a=0.025\ W.m^{-1}.K^{-1}$

inside temperature, $T_i=20^{\circ}C$

outside temperature, $T_o=-20^{\circ}C$

<u>From the Fourier's law of conduction we have the rate of heat transfer as:</u>

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=dT\div \frac{dx}{k.A}$

where:

k = thermal conductivity of the material

A = area subjected to the conduction

$dT =$ temperature difference across the two surfaces

dx = thickness of the surface

$\frac{dx}{k.A} =$ regarded as thermal resistance on electrical analogy

According to question here the heat transfer occurs due to conduction of the air.

Here we have two surfaces with air sandwiched between them. So we find an equivalent resistance:

$R_e=2\times \frac{x}{k.A}+ \frac{x'}{k_a.A}$

$R_e=2\times \frac{0.02}{0.78\times 1.5} +\frac{0.013}{0.025\times 1.5}$

$R_e=0.3808\ K.W^{-1}$

Therefore:

$\dot Q=40\div 0.3808$

$\dot Q=105.042\ W$

3 0

4 years ago

What is the speed of a vehicle that travels 80 meters in 2 seconds?

Firdavs [7]

the formula for velocity is:

v= distance/time

distance= 80m

time=2 seconds

v=80/2

v=40ms-1

8 0

3 years ago

A book sits on a bookshelf without moving until a student picks it up. Which law best explains why the book remains at rest unti

svetoff [14.1K]

Answer:

Newtons first law

Explanation:

object in rest stays at rest

object in motion stays in motion

8 0

3 years ago

The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy store

velikii [3]

Answer:

A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}$ (you has an mistake in the formula)

$\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}$

$\frac{1}{C_{eq1}}$ = 0.1 10⁶

$C_{eq1}$ = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

$\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}$

$\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6$

$\frac{1}{C_{eq2} }$ = 0.2 10⁶

$C_{eq2}$ = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

C_{eq} = C_{eq1} + C_{eq2}

C_{eq} = (10 + 5) 10⁻⁶

C_{eq} = 15 10⁻⁶ F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

Q = C_{eq} V

Q = 15 10⁻⁶ 6

Q = 90 10⁻⁶ C

the charge on each branch is

Q₁ = Ceq1 V

Q₁ = 10 10⁻⁶ 6

Q₁ = 60 10⁻⁶ C

Q₂ = C_{eq2} V

Q₂ = 5 10⁻⁶ 6

Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

In series combination the charge is constant Q = Q₁ = Q₃

U₃ = $\frac{Q^2}{2 C_3}$

U₃ = $\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}$

U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

U₄ = $\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}$

U₄ = 0.5 J

7 0

3 years ago

Pilings are driven into the ground at a building site by dropping a 2250 kg object onto them. What change in gravitational poten

den301095 [7]

Answer: 324.135 kJ

Explanation:

Given

Mass of dropping is $m=2250\ kg$

The initial height of dropping is $h_1=16\ m$

The final height of dropping $h_2=1.3\ m$

Gravitational potential energy is the function of height i.e.

$\Rightarrow \text{G.E.}=mgh$

Change in Gravitational Energy is

$\Rightarrow \Delta \text{G.E.}=mg(h_1-h_2)=2250\times 9.8\times (16-1.3)\\\\\Rightarrow \Delta \text{G.E.}=3,24,135\ J\approx 324.135\ kJ$

5 0

3 years ago