asinus

What is the square root of 98.38?

\(\sqrt{98.38}=9.91866926558 \)     !

x^2=27

\(x^2=27\)

\(x=\pm\sqrt{27}\)

\(x_1=5.19615242271\ \ x_2=-5.19615242271\)

  !

was ist 9-9a+?

\(Du \ kannst \ die \ 9 \ ausklammern.\)

\(9-9a=9\times (1-a)\)

  !

40*cos^2(x)=18*sin(x)+31 im Intervall 23pi-25pi

\(40\times cos^{2}x=18\times sin\ x+31\)

\(cos^2x=1-sin^2x\)

\(40\times(1-sin^2x)=18\times sin\ x+31\)      [s = sin x

\(40- 40s^2=18s+31\)

\(40s^2+18s -9=0\)

  a           b          c

\(s = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(s = {-18 \pm \sqrt{18^2-4\times 40\times (-9)} \over 2\times 40}\)

\(sin\ x=\frac{-18\pm\sqrt{1764}}{80}\)

\(sin\ x_1=0,3\)

\(x_1=arcsin\ 0,3=0,30469 (+24\pi\ 25\pi \ 26\pi)\)

\(sin\ x_2=-0,75 \)

\(x_2=arcsin(-0,75)=-0,84806(+24\pi\ 25\pi\ 26\pi)\)

     !

was ist größer 5/12 oder 9/16?

\(\frac{5}{12}=\frac{20}{48}<\frac{9}{16}=\frac{27}{48}\)   !

Frage: Welche Geschwindigkeit hat der Körper am Ende des 3m langen Weges?

Gegeben:

Steigung α = 55°

Masse m = 2 kg

Anfangsgeschwindigkeit V₀ = 0

Zugkraft F = 100 N

Weg s = 3 m

Reibungskraft F_R = 5 N

Gesamte Arbeit

\(W_g=F \times s=100N\times3m=\color{blue}300Nm\)

Reibungsarbeit

\(W_R= F_R \times s=5N\times 3m= \color{blue}15Nm\)

Hubarbeit

\(h = s\times sin55°=3m\times sin55°=\color{blue}2,457m\)

\(G=m\times g=2kg\times 9,81\frac{m}{s²}=\color{blue}19,62N\)

\(W_H=h\times G=2,457m\times 19,62N=\color{blue}48,206Nm\)

Bewegungsenergie \(E_B\)

Nach dem Satz von der Erhaltung der Energie ist

\(W_G=W_R+ W_H+E_B\)

\(E_B=W_G-W_R-W_H\)

\(E_B=300Nm-15Nm-48,206Nm=\color{blue}236,794Nm\)

Geschwindigkeit

\(\large E_B=\frac{m\times v²}{2}\)

\(\large v=\sqrt{\frac{2\times E_B}{m}}=\sqrt{\frac{2\times 236,794Nm}{2kg}}\) \(\times \sqrt{\frac{kg\times m}{N\times s²}}\)

\(\color{blue}v=15,388\frac{m}{s}\)       !

Hello mayazhanmath!

How much work is needed to lift an object that weighs 150 N to a height of 5 m?

\(The \ work \ done \ is \ 150N \times 5m.\)

\(W=F\times s = 150N\times5m = 750\ Nm \ or \ 750\ Ws.\)

\(The \ gravitational \ potential \ energy \)

\(grows \ equal \ this \ size.\)

Greeting asinus     ! Excuse my English, "around" is incorrect.

Hello mayazhanmath!

How much work is needed to lift an object that weighs 150 N to a height of 5 m?

\(The \ work \ done \ is \ 150N \times 5m.\)

\(W=F\times s = 150N\times5m = 750\ Nm \ or \ 750\ Ws.\)

\(The \ gravitational \ potential \ energy \)

\(grows \ around \ this \ size.\)

Greeting asinus     !

cos x = - sin x

\(\)

Greeting asinus    !

A custodianpulls a vacuum 13.5 m with a 33.9 N force at a 55.0 degree angle, against a 14.2 N friction force. Find the total work done on the vacuum. 

Please not so complicated.

The whole work is applied force times stretch.

\(W=F\times s = 33.9 N \times 13.5m=457.65 Nm\)

   !