Solve for Theda: sin2theda = 0.339

Thread starter mathcrzy
Start date Sep 20, 2007

AI Thread Summary

To solve for theta in the equation sin(2theta) = 0.339, first confirm that the equation represents the sine of double the angle, not the square of the sine. The right side has been calculated correctly as 0.339. To isolate theta, take the arcsin of 0.339, which gives you the angle for 2theta. Finally, divide the result by 2 to find theta.

Sep 20, 2007

mathcrzy

how do you solve for theda in the equation sin2theda=((2)(9.8))/(7.6^2).

i have figured the right side of the equation to be .339 and so I am left with sin2theda on the left and i don't know how to get theda by itself.

Physics news on Phys.org

Sep 20, 2007

nicksauce

Science Advisor

Homework Helper

Take the arcsin of the right side, then divide by 2.

Sep 21, 2007

jtbell

Staff Emeritus

Science Advisor

Homework Helper

mathcrzy said:

sin2theda

Do you mean \sin^2 \theta (square of the sine of theta) or \sin (2\theta) (sine of (two times theta))?

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »