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3 years ago

Which would be best categorized as heat transfer by convection?

Physics

1 answer:

pantera1 [17]3 years ago

6 0

Answer:

Explanation:

there are three modes of heat transfer which are

conduction
convection
radiation

Conduction is a mode of heat transfer in solid materials

Convection refers to a mode of heat transfer of fluids in motion.

Radiation occurs when heat moves as energy waves in form of infrared radiation. The energy from the sun reaches us by radiation.

From the list of options;

option A refers prevention of heat transfer by radiation

option B refers to noodles rising and falling in a poy of boiling water. The path which the noodles move shows the convective path of the fluid.

option C refers to heat transfer by conduction

Option D refers to oth convection and radiation as modes of heat transfer.

option B best describes heat transfer by convection

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A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?

dedylja [7]

Thew energy stored in a capacitor of capacitance $C$ and voltage between the plates $V$ is

$E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2$ .

Substituting numerical value

$E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J$

7 0

3 years ago

If a 6V battery is connected to a light bulb whose resistance is 55,000Ω How much current will flow in the circuit?

notka56 [123]

Answer:

Current, I = 0.000109 Amps

Explanation:

Given the following data;

Voltage = 6V

Resistance = 55,000 Ohms

To find the current flowing through the circuit;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

$V = IR$

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Making current the subject of formula, we have;

$I = \frac {V}{R}$

Substituting into the formula, we have;

$I = \frac {6}{55000}$

Current, I = 0.000109 Amps

5 0

3 years ago

The temperature of an air parcel and the kinetic energy of an air parcel are ___________ related. this means that as the tempera

torisob [31]

The temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.

<h3>What is temperature?</h3>

Temperature is the measure of degree of hotness or coldness of a body.

Temperature is also the measure of the average kinetic energy of a system.

When the heat is applied to body, its temperature increases as the body gains heat.

Thus, the temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.

Learn more about temperature here: brainly.com/question/25677592

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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

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

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