Solving √ x + 4/3 y with p = 3.4 e3 Kg m-³

  • Thread starter Thread starter stepha
  • Start date Start date
AI Thread Summary
To solve the equation √(x + (4/3)y) / p with given values p = 3.4 e3 Kg m-³, x = 1.6 e11 N m-², and y = 6.4 e10 N m-², substitute the values into the equation. It is recommended to use base units (kg, m, s) throughout the calculation to simplify unit conversions. After calculating the result, convert the final answer from m/s to km/s for the desired format. The discussion highlights the importance of unit consistency in solving the equation. Following these steps will lead to the correct solution.
stepha
Messages
8
Reaction score
0
Heres the question – :bugeye:

√ x + 4/3 y
√ p
(this whole equation should be squared, but not sure how to make the square root symbol big enough)

p = 3.4 e3 Kg m-³

x = 1.6 e11 N m-²

y = 6.4 e10 N m-²

I need to calculate the answer and give my answer in Km s-1 (Km per second)

Any ideas?? :confused:
 
Last edited:
Physics news on Phys.org
I don't understand... are you just trying to substitute them in?
 
If the problem is with the units, just use base units all the way through (kg, m, s) and then your final answer will be in m/s. Then its easy to convert m/s to km/s.
 
thank you for the advice, I will have another go
 
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...
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}...
Back
Top