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podryga [215]
3 years ago
8

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

Physics
2 answers:
ruslelena [56]3 years ago
7 0

Answer:

A

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
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Answer: B.

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Which of the following best defines the science of separating color wavelengths?
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Answer: spectroscopy

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What is Newton’s first law
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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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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&#10;\\\\&#10;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}&#10;

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)=&#10;\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}&#10;\\\\B_{in}=\frac{\mu_0kl^2}{3}&#10;

now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

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}&#10;

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)&#10;\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}&#10;\\\\B_{out}=\frac{\mu_0ka^3}{3r}&#10;

again using,equaiton 1,

B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}&#10;\\\\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}

So

           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
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