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3 years ago

Find the emitted power per square meter of peak intensity for a 3000 k object that emits thermal radiation.

2 answers:

8 0

According to Stefan-Boltzmann Law, the thermal energy radiated by a radiator per second per unit area is proportional to the fourth power of the absolute temperature. It is given by;
P/A = σ T⁴ j/m²s
Where; P is the power, A is the area in square Meters, T is temperature in kelvin and σ is the Stefan-Boltzmann constant, ( 5.67 × 10^-8 watt/m²K⁴)
Therefore;
Power/square meter = (5.67 × 10^-8) × (3000)⁴
= 4.59 × 10^6 Watts/square meter

steposvetlana [31]3 years ago

8 0

Explanation:

we can calculate total emitted power per square from Stefan-Boltzmann law

P = e σ AT⁴

where

P= power of body's thermal radiation

A= surface area of the body

σ= Stefan-Boltzmann constant it's value is $5.67 *10^-8 W/m^2k^4$

T= temperature of the body

e= emissivity of the substance here equal to 1

now by putting all the values in given formula

$P/A= (1) (5.67*10^-8 W/m^2K^4)(3000)^4$

we get the answer

$P/A= 4.5*10^ 6 W/m^2$

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Find the magnitude of acceleration (ft/s^2) a person experiences when he or she is texting and driving 58mph, hits a wall, and c

SVEN [57.7K]

Answer:

350 ft/s²

Explanation:

First, convert mph to ft/s.

58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s

Given:

v₀ = 85.1 ft/s

v = 0 ft/s

t = 0.24 s

Find: a

v = at + v₀

a = (v − v₀) / t

a = (0 ft/s − 85.1 ft/s) / 0.24 s

a = -354 ft/s²

Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².

7 0

3 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

Write the equation of a function h(t) that represents the amount of heat in joules required to heat the bar to a temperature of

bearhunter [10]

The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.

for 1 degree................... 7 Joules
y given degree........ p Joules

p=7y

In our case y=(t-25) .

h(t) = 7(t-25) which is the final answer.

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2 years ago

How can magnet be used to produce an electric current

Bess [88]

Answer:

Magnetic fields can be used to make electricity

Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current. Electricity generators essentially convert kinetic energy (the energy of motion) into electrical energy

Explanation:

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3 years ago

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hodyreva [135]

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