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Question Number 209432    Answers: 1   Comments: 0

A body is projected vertically upwards with a speed of 20m/s. Find the time in seconds when the body is 15m above it point of projection. g= 10m/s²

A body is projected vertically upwards with a speed of 20m/s. Find the time in seconds when the body is 15m above it point of projection. g= 10m/s²

Question Number 209430    Answers: 1   Comments: 0

Question Number 209320    Answers: 0   Comments: 0

Question Number 209193    Answers: 0   Comments: 1

A pin 6cm high is placed in front of a diverging lens of focal length 15cm, Calculate the position of the image formed

A pin 6cm high is placed in front of a diverging lens of focal length 15cm, Calculate the position of the image formed

Question Number 209129    Answers: 1   Comments: 0

Question Number 209030    Answers: 2   Comments: 0

Question Number 209027    Answers: 1   Comments: 0

Question Number 209026    Answers: 1   Comments: 0

Question Number 209023    Answers: 2   Comments: 1

Question Number 209016    Answers: 2   Comments: 0

Question Number 209015    Answers: 3   Comments: 0

Question Number 208975    Answers: 0   Comments: 0

Question Number 208970    Answers: 5   Comments: 0

Question Number 208969    Answers: 5   Comments: 0

Question Number 208968    Answers: 0   Comments: 0

hello everyone. Im writing a project on the topic“ Solution of Nonlinear partial differential equation using charpits methods” curently writing chapter 2 but I′m a little bit confused about what ih should include in my theoretical framework. Any ideas?

$$ \\ $$$$\mathrm{hello}\:\mathrm{everyone}.\:\mathrm{Im}\:\mathrm{writing}\:\mathrm{a}\:\mathrm{project}\:\:\mathrm{on}\:\mathrm{the}\:\mathrm{topic}``\:\mathrm{Solution}\:\mathrm{of} \\ $$$$\mathrm{Nonlinear}\:\mathrm{partial}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{using}\:\mathrm{charpits}\:\mathrm{methods}'' \\ $$$$\mathrm{curently}\:\mathrm{writing}\:\mathrm{chapter}\:\mathrm{2}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{a}\:\mathrm{little}\:\mathrm{bit}\:\mathrm{confused}\:\mathrm{about}\:\mathrm{what}\:\mathrm{ih} \\ $$$$\mathrm{should}\:\mathrm{include}\:\mathrm{in}\:\mathrm{my}\:\mathrm{theoretical}\:\mathrm{framework}.\:\mathrm{Any}\:\mathrm{ideas}? \\ $$

Question Number 208943    Answers: 1   Comments: 0

Question Number 208866    Answers: 1   Comments: 1

Question Number 208836    Answers: 2   Comments: 0

Question Number 208789    Answers: 0   Comments: 3

Why is surface tension formula divided by 2L that is, surface tension = (F/(2L)) why divided by 2L. Where did the 2 come from? Example. Calculate the force required to lift a needle 4cm long off the surface of water, if surface tension of water is 7.3 × 10^(−4) Nm^(− 1) Why is the formula surface tension = (F/(2L)) why not surface tension = (F/L)

$$\mathrm{Why}\:\mathrm{is}\:\mathrm{surface}\:\mathrm{tension}\:\mathrm{formula}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{2L} \\ $$$$\mathrm{that}\:\mathrm{is},\:\:\:\:\mathrm{surface}\:\mathrm{tension}\:\:=\:\:\frac{\mathrm{F}}{\mathrm{2L}} \\ $$$$\mathrm{why}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{2L}. \\ $$$$\mathrm{Where}\:\mathrm{did}\:\mathrm{the}\:\mathrm{2}\:\mathrm{come}\:\mathrm{from}? \\ $$$$ \\ $$$$\mathrm{Example}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{force}\:\mathrm{required}\:\mathrm{to}\:\mathrm{lift}\:\mathrm{a}\:\mathrm{needle} \\ $$$$\mathrm{4cm}\:\mathrm{long}\:\mathrm{off}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{water},\:\mathrm{if}\:\mathrm{surface} \\ $$$$\mathrm{tension}\:\mathrm{of}\:\mathrm{water}\:\mathrm{is}\:\:\mathrm{7}.\mathrm{3}\:×\:\mathrm{10}^{−\mathrm{4}} \mathrm{Nm}^{−\:\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{Why}\:\mathrm{is}\:\mathrm{the}\:\mathrm{formula}\:\:\:\mathrm{surface}\:\mathrm{tension}\:\:=\:\:\frac{\mathrm{F}}{\mathrm{2L}} \\ $$$$\mathrm{why}\:\mathrm{not}\:\:\:\:\:\:\mathrm{surface}\:\mathrm{tension}\:\:=\:\:\frac{\mathrm{F}}{\mathrm{L}} \\ $$

Question Number 208759    Answers: 0   Comments: 5

Question Number 208741    Answers: 1   Comments: 0

Question Number 208645    Answers: 1   Comments: 0

Solve : 2x_1 − λ_1 − 5λ_2 = 0 2x_2 − λ_1 − 2λ_2 = 0 2x_3 − 3λ_1 − λ_2 = 0 Find the values of x_1 , x_2 , x_3 , λ_1 , and λ_2

$$\mathrm{Solve}\:: \\ $$$$\mathrm{2x}_{\mathrm{1}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{5}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{2}} \:−\:\lambda_{\mathrm{1}} \:−\:\mathrm{2}\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{3}} \:−\:\mathrm{3}\lambda_{\mathrm{1}} \:−\:\lambda_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}_{\mathrm{1}} ,\:\mathrm{x}_{\mathrm{2}} ,\:\mathrm{x}_{\mathrm{3}} ,\:\lambda_{\mathrm{1}} ,\:\mathrm{and}\:\lambda_{\mathrm{2}} \\ $$

Question Number 208634    Answers: 0   Comments: 0

Question Number 208520    Answers: 1   Comments: 0

Question Number 208367    Answers: 0   Comments: 0

1. Find the length of each of the following (a) {x : −3 < x < 7} (b) {x : 2 ≤ x ≤ 6} ∪ {−3 ≤ x ≤ −1} (c) {x : −2 ≤ x < 5} ∪ {1 < x ≤ 7} 2. Let I=(a, b). Prove that I is measurable and m(I) = L(I).

$$\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\:\:\:\:\:\left(\mathrm{a}\right)\:\left\{\mathrm{x}\::\:−\mathrm{3}\:<\:\mathrm{x}\:<\:\mathrm{7}\right\} \\ $$$$\:\:\:\:\:\left(\mathrm{b}\right)\:\left\{\mathrm{x}\::\:\mathrm{2}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{6}\right\}\:\cup\:\left\{−\mathrm{3}\:\leqslant\:\mathrm{x}\:\leqslant\:−\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\left(\mathrm{c}\right)\:\left\{\mathrm{x}\::\:−\mathrm{2}\:\leqslant\:\mathrm{x}\:<\:\mathrm{5}\right\}\:\cup\:\left\{\mathrm{1}\:<\:\mathrm{x}\:\leqslant\:\mathrm{7}\right\} \\ $$$$ \\ $$$$\mathrm{2}.\:\mathrm{Let}\:\mathrm{I}=\left(\mathrm{a},\:\mathrm{b}\right).\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{I}\:\mathrm{is}\:\mathrm{measurable} \\ $$$$\mathrm{and}\:\mathrm{m}\left(\mathrm{I}\right)\:=\:\mathrm{L}\left(\mathrm{I}\right). \\ $$

Question Number 208312    Answers: 1   Comments: 0

lim_(x→0 ((a^x −1)/x) = log a)

$${lim}_{{x}\rightarrow\mathrm{0}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}\:=\:{log}\:{a}} \\ $$

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