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Dan1601
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Using SUVAT if s=(u+v)t/2
would t=2s/(u+v)?
would t=2s/(u+v)?
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SUVAT: Solving for t When s=(u+v)t/2
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To solve for time (t) using the equation s = (u + v)t/2, the correct rearrangement is t = 2s / (u + v). In the example of a car accelerating from 5 m/s to 20 m/s over 500 m, substituting the values gives t = 2 * 500 / (5 + 20), which simplifies to t = 40 seconds. To find acceleration (a), use the formula a = (v - u) / t, resulting in a = (20 - 5) / 40, which equals 0.375 m/s². The initial guess of 10 seconds was incorrect due to a miscalculation in the rearrangement.
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?
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}...
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...
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