Ask question

Ask question

All categories

2 years ago

12

You have a tungsten sphere (emissivity ε = 0.35) of radius 25 cm at a temperature of 25°C. If the sphere is enclosed in a room w

hose walls are kept at -5°C, what is the net flow rate of energy out of the sphere?

1 answer:

egoroff_w [7]2 years ago

8 0

Answer:

Explanation:

Stefan's formula for emission of radiation is

E = e σ A ( T⁴ - T₀⁴ )

E is energy radiated , e is emissivity , σ is stefan's constant , T is temperature of object and T₀ is temperature of surrounding. A is area of surface .

E = .35 x 5.67 x 10⁻⁸ ( 298⁴ - 268⁴ ) x 4π x .25²

= 1.9845 x 10⁻⁸ ( 78.86 - 51.58 ) x 10⁸ x .0625

= 3.38 J /s

You might be interested in

Wave-particle duality tells us that wave and particle models apply to all objects whatever the size, so why don't we observe wav

Genrish500 [490]

Answer:

Because the wavelengths of macroscopic objects are too short for them to be detectable.

Explanation:

Wavelength of an object is given by de Broglie wavelength as:

$\lambda=\frac{h}{mv}$

Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.

So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.

So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.

4 0

3 years ago

A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated to

Soloha48 [4]

To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.

The equation that describes the beat frequency is

$f_{beat} = |f_2-f_1|$

For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,

A) The frequency of the violin would be given by

$f_{beat} = |f_2-f_1|$

$5Hz = |f_2-440Hz|$

$f_2 = 440 \pm 5$

$f_2 = 445Hz or 435Hz$

B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.

5 0

3 years ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

7 0

3 years ago

The number of electromagnetic waves that travel past a certain point in a given time is the ________ of the radiation.

Rina8888 [55]

The period of the radiation?

8 0

3 years ago

Two blocks connected by a light string are being pulled across a frictionless horizontal tabletop by a hanging 16.2-N weight (bl

Artemon [7]

Newton's second law allows us to find the results for the string tensions are:

T₁ = 6.7 N

T₂ = 16.54 N

Newton's second law gives a relationship between force, mass and acceleration of bodies

∑ F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.

Free-body diagrams are representations of the forces applied to bodies without the details of them.

The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.

In the attachment we see a free-body diagram of the three-block system.

Let's apply Newton's second law to each body.

Block C

Y-axis

$W_c -T_2 = m_c a$

Block A

X axis

$T_2 - T_1 - W_a_x = m_a a$

Y axis

$N_a - W_a_y = 0$

Block B

X axis

$T_1 - W_b_x = m_b a$

Y axis

$N_b - W_b_y =0$

Let's use trigonometry to find the components of the weight.

Block A

cos θ = $\frac{W_a_y}{W_a}$

sin θ = $\frac{W_a_x}{W_a}$

$W_a_y = W_a cos \theta$

$W_a_x= W_a sin \theta$

Block B

cos θ = $\frac{W_b_y}{W_b}$

sin θ = $\frac{W_b_x}{W_b}$

$W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta$

Let's write our system of equations.

$W_c - T_2 = m_c a \\ T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a$

Let's find the acceleration of the bodies, adding the equations.

$W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a\\$

The weight is

W = mg

Let's substitute

$(m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b) \ a \\a= \frac{ m_c-m_a-m_b }{ m_a+m_b+m_c} \ g sin \theta$

Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.

m_c = Wc / g

m_c = 16.2 / 9.8

m_c = 1.65 kg

we substitute the values

$a= \frac{1.65 -2.20 -1.00}{1.65+2.20+1.00} \ 9.8 \ sin \theta \\a= -0.3096 sin \theta$

The negative sign indicates that the system is descending, to be able to give a specified value an angle is needed, they assume that the angle of the ramp is 45º

a = - 0.3196 sin 45

a = -0.226 m / s

Taking the acceleration we are going to look for the tensions.

From the equation of block C

$W_c - T_2 = m_c a \\T_2 = m_c ( g-a)\\T_2 = 1.65 ( 9.8 + 0.226)$

T₂ = 16.54 N

From the equation of block B

$T_1 - W_b_x = m_b a\\T_1 = m_b (a + g sin \theta)\\T_1 = 1.00 (-0.226 + 9.8 \ sin 45)$

T₁ = 6.7 N

In conclusion using Newton's second law we can find the results for the string tensions are:

T₁ = 6.7 N

T₂ = 16.54 N

Learn more here: brainly.com/question/20575355

7 0

3 years ago