Offspring get two alleles for each trait – one from each parent.
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
Reproducible is the term where your answers are similar to other groups. If every other group has similar results as yours (only if you use the same measurements) your results are reproducible. If your results are different to every other group your results aren't reproducible. <span />
Answer:
The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.
Explanation:
It is given that,
Mass of the box, m = 2.2 kg
The box is inclined at an angle of 30 degrees
Vertical distance, d = 3.1 m
The coefficient of friction, 
Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.


W is the work done by the friction.







v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.