Directly proportional to pressure
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
Answer:
900J
Explanation:
w =f×s
60×15
=900J
thus the k.e of the body is 900j
That was Tycho Brahe, and I thought it was actually more years than that.
Answer:
Explanation:
There will be reaction force by each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman
Taking torque of all forces about the end far away from the woman
Torque by reaction force = R₁ x 5.5
= 5.5 R₁ upwards
Torque by weight of woman in opposite direction , downwards
= - 804 x ( 5.5 - 1.55 )
= - 3175.8
Torque by weight of the plank in opposite direction , downwards .
= - 27 x 5.5 / 2
= - 74.25
Torque by R₂ will be zero as it passes through the point about which torque is being taken .
Total torque
= 5.5 R₁ - - 3175.8 - - 74.25 = 0 ( For equilibrium )
5.5 R₁ = 3250
R₁ = 590.9 N .
