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Physics-

General

Easy

Question

A hollow vertical cylinder of radius R is rotated with angular velocity w about an axis through its center. What is the minimum coefficient of static friction necessary to keep the mass M suspended on the inside of the cylinder as it rotates?

$μ = \frac{g R}{ω^{2}}$
$μ = \frac{ω^{2} g}{R}$
$μ = \frac{ω^{2} R}{g}$
$μ = \frac{g}{ω^{2} R}$

The correct answer is: $mu equals fraction numerator g over denominator omega to the power of 2 end exponent R end fraction$

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If $f left parenthesis x right parenthesis equals s i n to the power of negative 1 end exponent invisible function application x text and end text g left parenthesis x right parenthesis equals left square bracket s i n invisible function application x left parenthesis c o s invisible function application x right parenthesis right square bracket plus left square bracket c o s invisible function application left parenthesis s i n invisible function application x right parenthesis right square bracket comma text end text$ then range of f(g(x)) is

If $f left parenthesis x right parenthesis equals s i n to the power of negative 1 end exponent invisible function application x text and end text g left parenthesis x right parenthesis equals left square bracket s i n invisible function application x left parenthesis c o s invisible function application x right parenthesis right square bracket plus left square bracket c o s invisible function application left parenthesis s i n invisible function application x right parenthesis right square bracket comma text end text$ then range of f(g(x)) is

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If $f colon left parenthesis 0 comma pi right parenthesis rightwards arrow R$ defined by $f open parentheses x close parentheses equals stretchy sum from k equals 1 to n of open square brackets 1 plus sin invisible function application k x close square brackets comma$ where [x] denotes the integral part of x, then range of f(x) is

If $f colon left parenthesis 0 comma pi right parenthesis rightwards arrow R$ defined by $f open parentheses x close parentheses equals stretchy sum from k equals 1 to n of open square brackets 1 plus sin invisible function application k x close square brackets comma$ where [x] denotes the integral part of x, then range of f(x) is

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