Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate the power absorbed by any surface. The equation applies only to perfect radiators. The equation applies only to perfect absorbers. The equation is valid with any temperature units. The equation describes the transport of thermal energy by conduction.

Answers

Answer 1

Answer:

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

Explanation:

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.

$P=e \sigma A\left(T^(4)-T_(c)^(4)\right)$

So this law is application for calculating power absorbed by any surface.

Related Questions

(25) A grinding machine is supported on an isolator that has two springs, each with stiffness of k and one viscous damper with damping constant of c=1.8 kNs/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. The floor oscillates with amplitude Y=3 mm and frequency of 18 Hz. Because of other design constraints, the stiffness of each spring must be greater than 3.25 MN/m. What is the minimum required stiffness of each of the two springs to limit the grinding machine’s steady-state amplitude of oscillation to at most 10 mm? Assume that the grinding machine and the wheel are a rigid body of weight 4200 N and can move in only the vertical direction (the springs deflect the same amount).
1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?
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Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—than humans. What are the two lowest frequencies at which dogs have an increase in sensitivity? The speed of sound in the warm air of the ear is 350 m/s.A. 1700 Hz, 3400 Hz
B. 1700 Hz, 5100 Hz
C. 3400 Hz, 6800 Hz
D. 3400 Hz, 10,200 Hz

Answers

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. 1.11 × 104 Pa 1.09 × 105 Pa 1.13 × 107 Pa 1.11 × 108 Pa 2.18 × 105 Pa

Answers

Answer: 1.11 x 10⁸ Pa

Explanation:

At any deep, the absolute pressure is the same for all points located at the same level, and can be expressed as follows:

p = p₀ + δ. g . h, where p₀ = atmospheric pressure = 101, 325 Pa

Replacing by the values, we get:

p= 101,325 Pa + 1025 Kg/m³ . 9.8 m/s². 11,033 m = 1.11 x 10⁸ Pa.

Please give the answer

Answers

Answer:

Explanation:

just took it e2020

Answer:

60%

Explanation:

efficiency= useful/input x 100%,

Here, kinetic energy is useful for food processor (i.e. spinning blades)

600J/1000J=60%

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length l m 1 thickness t cm 0 8. and width w cm 4 and density . al kg m S 3 2820 a. Calculate the mass moment of inertia of the pendulum about its center of mass, . cm I b. Calculate the mass moment of inertia of the pendulum about its pivot point, o

Answers

Answer:

1) I_ pendulum = 2.3159 kg m², 2) I_pendulum = 24.683 kg m²

Explanation:

In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum

Sphere

They indicate the density of the sphere roh = 37800 kg / m³ and its radius

r = 5 cm = 0.05 m

we use the definition of density

ρ = M / V

M = ρ V

the volume of a sphere is

V = 4/3 π r³

we substitute

M = ρ 4/3 π r³

we calculate

M = 37800 4/3 π 0.05³

M = 19,792 kg

Bar

the density is ρ = 32800 kg / m³ and its dimensions are 1 m,

0.8 cm = 0.0008 m and 4cm = 0.04 m

The volume of the bar is

V = l w h

m = ρ l w h

we calculate

m = 32800 (1 0.008 0.04)

m = 10.496 kg

Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is

r_cm = 1 / M (M r_sphere + m r_bar)

M = 19,792 + 10,496 = 30,288 kg

r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))

r_cm = 1 / 30,288 (5,248 + 20,7816)

r_cm = 0.859 m

This is the center of mass of the pendulum.

1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:

Sphere I = 2/5 M r2

Bar I = 1/12 m L2

parallel axes theorem

I = I_cm + m D²

where m is the mass of the body and D is the distance from the body to the axis of rotation

Sphere

m = 19,792 ka

the distance D is

D = 1.05 -0.85

D = 0.2 m

we calculate

I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168

I_sphere = 0.811472 kg m²

Bar

m = 10.496 kg

distance D

D = 0.85 - 0.5

D = 0.35 m

I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576

I_bar = 1.5044 kg m²

The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts

I_pendulum = I_sphere + I_bar

I_pendulum = 0.811472 +1.5044

I_ pendulum = 2.3159 kg m²

this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m

2) The moment is requested with respect to the pivot point at r = 0 m

Sphere

D = 1.05 m

I_sphere = 2/5 M r2 + M D2

I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82

I_sphere = 21.84 kg m²

Bar

D = 0.5 m

I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624

I_bar = 2,84266 kg m 2

The pendulum moment of inertia is

I_pendulum = 21.84 +2.843

I_pendulum = 24.683 kg m²

This moment of inertia is about the turning point at r = 0 m

A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?

Answers

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

$v^2=u^2+2as$

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

$0^2=u^2-2* 9.81* 0.76\n\n\therefore u=√(2* 9.81* 0.76)=3.86m/s$

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

$v^(2)=3.86^2-2* 9.81* 0.66\n\n\therefore v=√(3.86^2-2* 9.81* 0.66)=1.3966m/s$

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

$v=u+at$

where symbols have the usual meaning

Applying the given values we get

$t=(v-u)/(g)\n\nt=(0-1.3966)/(-9.81)=0.1423seconds$

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