Science, asked by anshikapjs2891, 5 months ago

Which may result from an increase in friction? decreased traction increased speed reduced wear and tear generation of heat

Answers

Answered by Anonymous
3

Answer:

\huge\bold{Question :}

If tan θ = ¹/√7 then , show that \sf \dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}=\dfrac{3}{4}

\huge\bold{Solution :}

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\sf tan\ \theta=\dfrac{1}{\sqrt{7}}

:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}

:\to \sf \textsf{\textbf{\pink{tan$^\text{2} \boldsymbol \theta\ $ =\ $\dfrac{\text{1}}{\text{7}}$}}}\ \; \bigstar

\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}

:\to \sf cot\ \theta=\sqrt{7}

:\to \sf cot^2\theta=(\sqrt{7})^2

:\to \sf \textsf{\textbf{\green{cot$^\text{2}\ \boldsymbol \theta $\ =\ 7}}}\ \; \bigstar

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LHS

:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}

From Trigonometric identities ,

csc²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}

tan²θ = ¹/₇

cot²θ = 7

:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}

:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}

:\to\ \sf \dfrac{48}{64}

:\to\ \textsf{\textbf{\orange{$\dfrac{\text{3}}{\text{4}}$}}}\ \; \bigstar

Answered by stalwartajk
1

Answer: The friction can be increased by creating a rougher surface , by pressing the two surfaces with greater force and by removing the lubricants from two surfaces.

Explanation: The following methods by which The friction can be increased are-

  1. by creating a rougher surface
  2. by pressing the two surfaces with greater force
  3.  by removing the lubricants from two surfaces.
  4. by using sliding motion instead of rolling motion.

The following methods by which The friction can be decreased are-

  1. by using the lubricants.
  2. by polishing the surface.
  3. for reducing fluid friction ships, boats and airplanes are designed as streamlined .
  4. ball bearings are used in rotatory machines to reduce friction .

To know more about friction from the given link

https://brainly.in/question/457850.

To know more about motion from the given link

https://brainly.in/question/18408438

#SPJ3

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