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3 years ago

IF YOU GET GOOD GRADES ON SCIENCE TESTS PLEASE HELP ME!!!!!

Physics

2 answers:

ruslelena [56]3 years ago

7 0

Answer:

Explanation:

The sound wave hits the apartment complex first.

im sure this is physics and not science

Elena-2011 [213]3 years ago

5 0

Answer:

A i think hope this helps

Explanation:

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Three polarizing filters are shown in the figure below. The alignment of each filterâs transmission axis is shown, as measured f

lidiya [134]

Answer: B.

Explanation:

Hope help :p

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3 years ago

Which of the following best defines the science of separating color wavelengths?

Mice21 [21]

Answer: spectroscopy

Spectroscopy is the separation of the light in the different wavelengths and spectrophotometry measures the intensities of the different components of the light to get the composition of substances.

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3 years ago

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What is Newton’s first law

amid [387]

Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

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3 years ago

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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

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in

iVinArrow [24]

Answer:

The value is $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 14\ psla$

The initial temperature is $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

The final temperature is $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

$PT^{\frac{\gamma}{1- \gamma} } = Constant$

=> $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas $\gamma = \frac{5}{3}$

$14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=> $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=> $P_2 = 40.54 \ psla$

8 0

2 years ago