Question

a battery circulates a charge around a circuit of one minute. if the current in the circuit is 5 ampheres, what quantity of charge passes throughhh the battery.​

Answer 1

$\underline{\large{\sf\ \ \ \ \ Given:-\ \ \ \ }}$

• • Time, t = 1 min or, 60 sec
• • Current, l = 5 A

$\underline{\large{\sf\ \ \ \ \ To \: Find:-\ \ \ \ }}$

• • What's the quantity of charge flow through the battery ?

$\underline{\large{\sf\ \ \ \ \ SOLUTION:-\ \ \ \ }}$

$\underline{\:\textsf{ We know that :}}$

$\boxed{\sf\ I= \dfrac{Q}{t}}$

$\underline{\:\textsf{ Putting the values :}}$

$\begin{gathered} \dashrightarrow \sf Q= I\times t\\ \\\ \dashrightarrow \sf Q= 5\times 60\\ \\ \dashrightarrow \sf Q= 300C\end{gathered}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\begin{gathered} \therefore \underline{ \textsf{ Amount of charge passes 300 C}} \\ \end{gathered}$

Answer 2

Answer:

$\huge{\red{\underline{\overline{ᴀ}}}}{\color{orange}{\underline{\overline{ɴ}}}}{\color{gold}{\underline{\overline{s}}}}{\color{green}{\underline{\overline{ᴡ}}}}{\blue{\underline{\overline{ᴇ}}}}{\pink{\underline{\overline{ʀ}}}}\star$

$\odot \: \: \: \: \: \: \red{ \underline{ \underline{ \green{ \bf{Length \: of \: MN \: is \: 21cm.}}}}}$

ㅤ

Step-by-step explanation:

Given :

$\sf where \begin{cases} & \sf{length \: of \: AB = 24cm} \\ \\ & \sf{ length \: of \: CD = 18cm} \\ \\ & \sf{radius \: of \: the \: circle = 15cm} \end{cases}$

To Find :

the length of MN

Solution :

$\underline{ \frak{ \dag \: as \: we \: know}}$

$\boxed{ \pink{ \sf{Pythagoras \: theorem \implies \: AC² = AB² + BC²}}}$

__________________________________________

The ∆CNO is a right angle triangle

And N is the mid-point of CD

$\\ \therefore \sf \frac{1}{2} CD=CN$

$\\ \sf \to \frac{1}{2} \times 18 =CN$

$\\ \sf \to \: CN = 9cm$

By using Pythagoras theorem

$\\ \dashrightarrow \sf \: CO² = CN² + NO²$

$\\ \dashrightarrow \sf \: {15}^{2} = {9}^{2} + NO²$

$\\ \dashrightarrow \sf \: 225 = 81 + NO²$

$\\ \dashrightarrow \sf \: NO² = 225 - 81$

$\\ \dashrightarrow \sf \: NO² = 144$

$\\ \dashrightarrow \sf \: NO = \sqrt{144}$

$\\ \dashrightarrow \sf \: NO = 12$

Therefore, the Length of NO is 12cm

ㅤ

The ∆AOM is a right angle triangle

And M is the mid-point of AB

$\\ \therefore \sf \frac{1}{2} AB = AM$

$\\ \sf \to \frac{1}{2} \times 24 = AM$

$\\ \sf \to \: AM = 12cm$

By Using Pythagoras Theorem

$\\ \sf \dashrightarrow \: OA² = OM² + MA²$

$\\ \dashrightarrow \sf \: {15}^{2} = OM² + {12}^{2}$

$\\ \dashrightarrow \sf \: 225 = OM² + 144$

$\\ \dashrightarrow \sf \: OM² = 225 - 144$

$\\ \dashrightarrow \sf \: OM² = 81$

$\\ \dashrightarrow \sf \: OM = \sqrt{81}$

$\\ \dashrightarrow \sf \: OM = 9$

Therefore, the length of OM is 9cm

ㅤ

Therefore, the Length of MN

= Length of ON + Length of OM

$= 12 + 9$

$= \large{\underline{ \boxed{ \frak{ \purple{21cm}}}} \star}$

__________________________________________

$\underline{ \text{Therefore, the Length of MN is 21cm.}}$

Answer 3

Answer:

$\huge{\red{\underline{\overline{ᴀ}}}}{\color{orange}{\underline{\overline{ɴ}}}}{\color{gold}{\underline{\overline{s}}}}{\color{green}{\underline{\overline{ᴡ}}}}{\blue{\underline{\overline{ᴇ}}}}{\pink{\underline{\overline{ʀ}}}}\star$

$\odot \: \: \: \: \: \: \red{ \underline{ \underline{ \green{ \bf{Length \: of \: MN \: is \: 21cm.}}}}}$

ㅤ

Step-by-step explanation:

Given :

$\sf where \begin{cases} & \sf{length \: of \: AB = 24cm} \\ \\ & \sf{ length \: of \: CD = 18cm} \\ \\ & \sf{radius \: of \: the \: circle = 15cm} \end{cases}$

To Find :

the length of MN

Solution :

$\underline{ \frak{ \dag \: as \: we \: know}}$

$\boxed{ \pink{ \sf{Pythagoras \: theorem \implies \: AC² = AB² + BC²}}}$

__________________________________________

The ∆CNO is a right angle triangle

And N is the mid-point of CD

$\\ \therefore \sf \frac{1}{2} CD=CN$

$\\ \sf \to \frac{1}{2} \times 18 =CN$

$\\ \sf \to \: CN = 9cm$

By using Pythagoras theorem

$\\ \dashrightarrow \sf \: CO² = CN² + NO²$

$\\ \dashrightarrow \sf \: {15}^{2} = {9}^{2} + NO²$

$\\ \dashrightarrow \sf \: 225 = 81 + NO²$

$\\ \dashrightarrow \sf \: NO² = 225 - 81$

$\\ \dashrightarrow \sf \: NO² = 144$

$\\ \dashrightarrow \sf \: NO = \sqrt{144}$

$\\ \dashrightarrow \sf \: NO = 12$

Therefore, the Length of NO is 12cm

ㅤ

The ∆AOM is a right angle triangle

And M is the mid-point of AB

$\\ \therefore \sf \frac{1}{2} AB = AM$

$\\ \sf \to \frac{1}{2} \times 24 = AM$

$\\ \sf \to \: AM = 12cm$

By Using Pythagoras Theorem

$\\ \sf \dashrightarrow \: OA² = OM² + MA²$

$\\ \dashrightarrow \sf \: {15}^{2} = OM² + {12}^{2}$

$\\ \dashrightarrow \sf \: 225 = OM² + 144$

$\\ \dashrightarrow \sf \: OM² = 225 - 144$

$\\ \dashrightarrow \sf \: OM² = 81$

$\\ \dashrightarrow \sf \: OM = \sqrt{81}$

$\\ \dashrightarrow \sf \: OM = 9$

Therefore, the length of OM is 9cm

ㅤ

Therefore, the Length of MN

= Length of ON + Length of OM

$= 12 + 9$

$= \large{\underline{ \boxed{ \frak{ \purple{21cm}}}} \star}$

__________________________________________

$\underline{ \text{Therefore, the Length of MN is 21cm.}}$