Question :- If a spring of spring constant ‘m’ is divided into 5:3 ratio, then what will be the spring constant of the larger piece of spring?

Answer :- 8m/5

I need explanation! ​

Answers

Answer 1

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

Here we go ~

As we have been given, The spring of spring constant " m " was cut to form 2 new springs in ratio of 5 : 3.

we have to find out the spring constant of the longer spring which was formed, and it's given that spring constant of the main spring was " m "

Now, let the length of main spring be " l "

So, lengths of the resultant springs will be :

[tex]\qquad \cal \dashrightarrow \: \dfrac{5l}{8} \: \: and \: \: \dfrac{3l}{8} \: \: \: \: respectively[/tex]

And we know that spring constant is inversely proportional to length of spring so, we infer that :

[tex]\qquad \cal \dashrightarrow \: m\propto \dfrac{1}{l} m \: \: \: \: \: - (1)[/tex]

where, m is spring constant of main spring and l is its length.

[tex]\qquad \cal\dashrightarrow \: m_1 \propto \dfrac{8}{5l} \: \: \: \: \: \: \: - (2)[/tex]

where, m1 is spring constant of longer spring, and 5l/8 is its length.

let's divide (2) by (1), we will get ~

[tex]\qquad \cal \dashrightarrow \: \dfrac{m_1}{m} = \dfrac{8}{5l} \div \dfrac{1}{l} [/tex]

[tex]\qquad \cal \dashrightarrow \: \dfrac{m_1}{m} = \dfrac{8}{5l} \times \dfrac{l}{1} [/tex]

[tex]\qquad \cal \dashrightarrow \: \dfrac{m_1}{m} = \dfrac{8}{5} [/tex]

[tex]\qquad \cal \dashrightarrow \: {m_1}{} = \dfrac{8}{5} m [/tex]

So, the spring constant of equivalent longer strong formed will be 8/5 m

Answer 2

prove:

[tex]m_2 = \frac{8m}{5} [/tex]

Explanation:

Given:

Spring constant of spring= m

The ratio of spring = 5:3

To find:

spring constant of the larger piece of spring?

Solution:

Let the length of larger piece L2 be 5x and smaller piece L1 be 3x with spring constant m2 and m1 respectively.

now,

spring constant (m) inversely proportional to the length of spring i.e.

[tex]m \propto \: \frac{1}{l} {,}\: m_1 \propto \: \frac{1}{l_1} {,}\: m_2 \propto \: \frac{1}{l_2}[/tex]

Also,

The spring is connected in series hence,

[tex] \frac{1}{m} = \frac{1}{m_1} + \frac{1}{m_2} [/tex]

and

[tex] \frac{m_1}{m_2} = \frac{l_2}{l_1} \\ {m_1} = {m_2} \cdot \frac{l_2}{l_1} \\ {m_1} = {m_2} \cdot \frac{5}{3} \\ [/tex]

Substituting above value in,

[tex] \frac{1}{m} = \frac{1}{m_1} + \frac{1}{m_2} \\ \frac{1}{m} = \frac{1}{{m_2} \cdot \frac{5}{3}} + \frac{1}{m_2} \\ \frac{1}{m} = \frac{3}{5m_2} + \frac{1}{m_2} \\ \frac{1}{m} = \frac{3m_2 + 5m_2}{5 {m_2}^{2} } \\ \frac{1}{m} = \frac{8m_2}{5 {m_2}^{2} } \\ \frac{1}{m} = \frac{8 \cancel {m_2}}{5 \cancel{m_2}^{2} } \\ \frac{1}{m} = \frac{8}{5m_2} \\ m_2 = \frac{8m}{5} [/tex]

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An electronic device has a resistance of 20 ohms and a current of 15 A. What is the voltage across the device ?​

Answers

Question :-

An Electronic Device has a Resistance of 20 Ohm's and a Current of 15 Ampere . What is the Voltage across the Device ?

Answer :-

Voltage of the Device is 300 Volt's .

[tex] \rule {180pt}{2pt} [/tex]

Given :-

Resistance = 20 Ohm'sCurrent = 15 Ampere

To Find :-

Voltage = ?

Solution :-

As per the provided information in the given question, we have been given that the Resistance of the Device is 20 Ohm's. It's Current is given as 15 Ampere . And, we have been asked to calculate the Voltage .

For calculating the Voltage , we will use the Formula :-

[tex] \bigstar \: \: \: \boxed {\sf { \: Voltage \: = \: Current \: \times \: Resistance \: }} [/tex]

Therefore , by Substituting the given values in the above Formula :-

[tex] \Longrightarrow \: \: \: \sf { Voltage \: = \: Current \: \times \: Resistance } [/tex]

[tex] \Longrightarrow \: \: \: \sf { Voltage \: = \: 15 \: \times \: 20 } [/tex]

[tex] \Longrightarrow \: \: \: \bf { Voltage \: = \: 300 } [/tex]

Hence :-

Voltage of Device = 300 Volt's .

[tex] \underline {\rule {180pt}{4pt}} [/tex]

Additional Information :-

[tex] \Longrightarrow \: \: \: \sf {Voltage \: = \: Current \: \times \: Resistance} [/tex]

[tex] \Longrightarrow \: \: \: \sf {Current \: = \: \dfrac {Voltage}{Resistance}} [/tex]

[tex] \Longrightarrow \: \: \: \sf {Resistance \: = \: \dfrac {Voltage}{Current} } [/tex]

Answer:

300 volts

Explanation:

now in this question we have been given our resistance which is 20 ohms and our current which is 15A

using ohm's law

V=IR

where v is voltage, I is current and r is resistance

now substitute,

V=15 x 20

V=300

I hope I helped

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Answer:

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B. The protons flow through the load.
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Answers

The function of the load in an electric circuit is it safely opens and closes the circuit. Thus, the correct option is D.

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I got it right on edge 2022

Answer:

c

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