Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1558

Question Number 47967    Answers: 1   Comments: 0

A particle moves in a linear scare such that acceleration after t seconds is a ms^(−2) where a= 2t^2 + t.If its initial velocity was 3ms^(−1) find an expression for S,the distance in meters traveled from start t seconds.

$${A}\:{particle}\:{moves}\:{in}\:{a}\:{linear}\:{scare}\:{such}\:{that}\:{acceleration} \\ $$$${after}\:{t}\:{seconds}\:{is}\:{a}\:{ms}^{−\mathrm{2}} \:{where}\:{a}=\:\mathrm{2}{t}^{\mathrm{2}} +\:{t}.{If}\:{its}\:{initial}\: \\ $$$${velocity}\:{was}\:\mathrm{3}{ms}^{−\mathrm{1}} \:{find}\:{an}\:{expression}\:{for}\:{S},{the}\:{distance}\:{in}\:{meters} \\ $$$${traveled}\:{from}\:{start}\:{t}\:{seconds}. \\ $$

Question Number 47966    Answers: 1   Comments: 1

a(x^2 +y^2 )+b(x+y)= c & x^2 −y^2 = R^2 Solve for x or y .

$${a}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+{b}\left({x}+{y}\right)=\:{c} \\ $$$$\:\&\:\:\:\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} \\ $$$${Solve}\:{for}\:{x}\:{or}\:{y}\:. \\ $$

Question Number 47965    Answers: 0   Comments: 0

A small body is made to travel linearly t sconds after the start.Its distance S meters from a fix point O on a linear scale is given by S = t^2 −5t + 6. a) How far is the body from O at the start? b)with what velocity does it start? c)when is the body momentarily at rest? d) What is the acceleration of the body?

$${A}\:{small}\:{body}\:{is}\:{made}\:{to}\:{travel}\:{linearly}\:{t}\:{sconds}\:{after}\:{the}\: \\ $$$${start}.{Its}\:{distance}\:{S}\:{meters}\:{from}\:{a}\:{fix}\:{point}\:{O}\:{on}\:{a}\: \\ $$$${linear}\:{scale}\:{is}\:{given}\:{by}\:{S}\:=\:{t}^{\mathrm{2}} −\mathrm{5}{t}\:+\:\mathrm{6}. \\ $$$$\left.{a}\right)\:{How}\:{far}\:{is}\:{the}\:{body}\:{from}\:{O}\:{at}\:{the}\:{start}? \\ $$$$\left.{b}\right){with}\:{what}\:{velocity}\:{does}\:{it}\:{start}? \\ $$$$\left.{c}\right){when}\:{is}\:{the}\:{body}\:{momentarily}\:{at}\:{rest}? \\ $$$$\left.{d}\right)\:{What}\:{is}\:{the}\:{acceleration}\:{of}\:{the}\:{body}? \\ $$

Question Number 47964    Answers: 0   Comments: 0

A particle A at rest with position vector 2i−j with mass 500kg is hit by another particle B moving at a velocity of (5i−4j)ms^(−1) with mass 300kg. and they all move in the direction of B. a) What is the momentum after impact? b) Calculate the impulse generated,hence or otherwise, Calculate the distance A and B cover after impact at a time of 1minute.

$${A}\:{particle}\:{A}\:{at}\:{rest}\:{with}\:{position}\:{vector}\:\mathrm{2}{i}−{j}\:\:{with}\:{mass}\:\mathrm{500}{kg}\:{is}\:{hit}\:{by}\: \\ $$$${another}\:{particle}\:{B}\:{moving}\:{at}\:{a}\:{velocity}\:{of}\:\left(\mathrm{5}{i}−\mathrm{4}{j}\right){ms}^{−\mathrm{1}} \:{with}\:{mass}\:\mathrm{300}{kg}. \\ $$$${and}\:{they}\:{all}\:{move}\:{in}\:{the}\:{direction}\:{of}\:{B}. \\ $$$$\left.{a}\right)\:{What}\:{is}\:{the}\:{momentum}\:{after}\:{impact}? \\ $$$$\left.{b}\right)\:{Calculate}\:{the}\:{impulse}\:{generated},{hence}\:{or}\:{otherwise}, \\ $$$${Calculate}\:{the}\:{distance}\:{A}\:{and}\:{B}\:{cover}\:{after}\:{impact}\:{at}\: \\ $$$${a}\:{time}\:{of}\:\mathrm{1}{minute}. \\ $$$$ \\ $$

Question Number 47960    Answers: 1   Comments: 0

Question Number 47979    Answers: 0   Comments: 8

Question Number 47957    Answers: 3   Comments: 3

Question Number 47947    Answers: 0   Comments: 1

Question Number 47939    Answers: 1   Comments: 2

Question Number 47938    Answers: 0   Comments: 1

Question Number 47932    Answers: 1   Comments: 1

Question Number 47917    Answers: 0   Comments: 0

Question Number 47916    Answers: 2   Comments: 1

Question Number 47915    Answers: 2   Comments: 0

Question Number 47912    Answers: 0   Comments: 0

Question Number 47908    Answers: 0   Comments: 0

Question Number 47906    Answers: 0   Comments: 2

Question Number 47903    Answers: 0   Comments: 6

Question Number 47902    Answers: 0   Comments: 1

Question Number 47900    Answers: 0   Comments: 0

thanks sir

$${thanks}\:{sir} \\ $$

Question Number 47899    Answers: 0   Comments: 0

Question Number 47896    Answers: 1   Comments: 0

3(5/7)/2(1/7)=×/1.5 sir plz help me

$$\mathrm{3}\frac{\mathrm{5}}{\mathrm{7}}/\mathrm{2}\frac{\mathrm{1}}{\mathrm{7}}=×/\mathrm{1}.\mathrm{5}\:\:\:\:\:\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$$$ \\ $$

Question Number 47883    Answers: 2   Comments: 1

Question Number 47881    Answers: 0   Comments: 1

Question Number 47874    Answers: 0   Comments: 0

An elevator starts with m passengers and stops at n floors (m≤n). The probability that no passengers alight at the same floor is

$$\mathrm{An}\:\mathrm{elevator}\:\mathrm{starts}\:\mathrm{with}\:{m}\:\mathrm{passengers} \\ $$$$\mathrm{and}\:\mathrm{stops}\:\mathrm{at}\:{n}\:\mathrm{floors}\:\left({m}\leqslant{n}\right).\:\mathrm{The}\: \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{no}\:\mathrm{passengers}\:\mathrm{alight} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{floor}\:\mathrm{is} \\ $$

Question Number 47863    Answers: 0   Comments: 0

find ∫ (√(1−x^4 ))dx

$${find}\:\int\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$

  Pg 1553      Pg 1554      Pg 1555      Pg 1556      Pg 1557      Pg 1558      Pg 1559      Pg 1560      Pg 1561      Pg 1562   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com