Question

if polynomial 6x⁴ + 8x³+ 17x² +21x + 5 is divided by another polynomial 3x² + 4x + 1 remainder comes out to be ax + b. find a and b​

Answer 1

Answer:

Given that , A Solenoid with turns per cm carries a current of 1 A [ or , Ampere ] .

Exigency To Find : The magnitude energy stored per unit volume ?

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⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⌬⠀Current , I = 1 A [ or Ampere ]

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⌬ ⠀Number of turns per unit length , n = 40 turns per m .

⠀⠀⠀⠀¤ Converting the Number of turns from per m ( metre ) to per cm ( centimeter ) :

\begin{gathered} \qquad:\implies \sf Number \:_{\:(\:Turns \: )\:} \:=\: 40 \:turns \:per \: m \:\:\\\\ \qquad:\implies \sf Number \:_{\:(\:Turns \: )\:} \:=\: 40 \times 100 \:turns \:per \: cm \:\:\qquad \because \bigg\lgroup \sf{ 1m \: =\: 100 \:cm}\bigg\rgroup\\\\ \qquad:\implies \underline {\boxed{\pmb{\frak{ Number \:_{\:(\:Turns \: )\:} \:=\: 4000 \:turns \:per \: cm\:}}}} \:\:\bigstar \\\\ \end{gathered}:⟹Number(Turns)=40turnsperm:⟹Number(Turns)=40×100turnspercm∵⎩⎪⎪⎪⎧1m=100cm⎭⎪⎪⎪⎫:⟹Number(Turns)=4000turnspercmNumber(Turns)=4000turnspercm★

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding Energy stored per unit Volume :

As , We know that ,

⠀⠀⠀⠀⠀⠀⠀ENERGY STORED PER UNIT VOLUME , [ u ] :

\begin{gathered}\qquad \star \:\:\underline {\boxed {{\pmb{\sf{  \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{ 1} {2\mu _0 \:} \:\:\times \:B^2}\bigg\rgroup  \:\:Jm^{-3}\:}}}}}\\\\\end{gathered}⋆ u=⎩⎪⎪⎪⎧2μ01×B2⎭⎪⎪⎪⎫ Jm−3 u=⎩⎪⎪⎪⎧2μ01×B2⎭⎪⎪⎪⎫ Jm−3

\begin{gathered}\qquad \dashrightarrow \sf \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{ 1} {2\mu _0 \:} \:\:\times \:B^2}\bigg\rgroup  \:\:Jm^{-3}\: \\\\ \qquad \dashrightarrow \sf \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{ 1}{ 2\mu _0 \:} \:\:\times \:\{ \mu_0 n I \}^2 }\bigg\rgroup  \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{(\:\mu_0 n I\: )^2} {2\mu _0 \:} \: }\bigg\rgroup  \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \dfrac{(\:\mu_0 n I\: )^2} {2\mu _0 \:} \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \dfrac{\:\mu_0 n^2 I^2} {2 \:} \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \dfrac{\:4 \pi \:\times 10^{-7} \times ( 4000 )^2 \times 1^2} {2 \:} \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \:2 \pi \:\times 10^{-7} \times ( 4000 )^2 \times 1\: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \:32 \pi \:\times 10^{-1} \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \:3.2 \pi \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\frak{ Energy \:Stored \:\:(\:or \;u\:)\:=\:\:3.2 \pi \: \:\:Jm^{-3} \:}}}} \:\:\bigstar \\\\ \end{gathered}⇢u=⎩⎪⎪⎪⎧2μ01×B2⎭⎪⎪⎪⎫ Jm−3⇢u=⎩⎪⎪⎪⎧2μ01×{μ0nI}2⎭⎪⎪⎪⎫ Jm−3⇢u=⎩⎪⎪⎪⎧2μ0(μ0nI)2⎭⎪⎪⎪⎫ Jm−3⇢u=2μ0(μ0nI)2Jm−3⇢u=2μ0n2I2Jm−3⇢u=24π×10−7×(4000)2×12Jm−3⇢u=

Answer 2

Step-by-step explanation:

Heat produce - Electricity

Before we solve the Question to get the quantity of heat produced, let's first know that how to solve this question.

What we need to do

Here in this question we are given with current passed, resistance and time and we are unknown with the heat produce, so we need to calculate the quantity of heat produce.

To find the quantity of heat produce firstly we need to convert the time from minutes into seconds.

In order to convert the time from minutes into second, we need to multiply the time value by 60. Because we know There are 60 seconds in 1 minute.

After that by using the required formula of heat we can easily calculate the quantity of heat produced. We then solve the problem and we understand the steps to get our final answer of heat produced.

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Required Solution

Given that the current of 2A is passed through a resistance of 250 ohms for 5 min. And we are asked to calculate the quantity of heat produced.

Given values in the Question,

– Current = 2A

– Resistance = 250 Ω

– Time = 5 min

– Heat produced = Unknown.

Let's first convert the conversation of time from minutes to seconds.

We know that to convert the time from minutes into second, we need to multiply the time value by 60.

➝ Time = 5

➝ Time = 5 × 60

➝ Time = 300

Therefore, the time in second is 300 sec.

Now we know that,

\tt{\dashrightarrow\boxed{\tt{H = {(I)}^{2} \times Rt}}}⇢H=(I)2×Rt

Here, H denotes heat produced, I denotes Current passed, R denotes resistance and t denotes time.

Now by using the required formula and Substituting all the given values in the formula, we get :

\tt{:\implies H = (2)^2 \times 250 \times 300}:⟹H=(2)2×250×300

\tt{:\implies H = 4 \times 250 \times 300}:⟹H=4×250×300

\tt{:\implies H = 1000 \times 300}:⟹H=1000×300

\tt{:\implies \boxed{\boxed{\tt{H = 300000}}}}:⟹H=300000

Hence, the quantity of heat and is 300000 Joule.