Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
1/1000 of a meter.
The prefix milli- means 1/1000.
S= 1/2 x 182 x t = 1688.3
t = 1688.3 / 91
time = 18.55 seconds
Answer:
increases
Explanation:
when the mass of an object is constant and you add more force, acceleration increases
We study the motion of the ball on the two coordinating needles Ox and Oy, originated on the ground. On Ox the ball has a uniform straight motion with the velocity vx = vo and the equation of motion is x = vo * t, and on the ball Oy has a movement uniformly accelerated with acceleration g and no initial velocity on the axis Oy v_oy = 0 (free fall), so the equation of motion is y = h-gt² / 2.The equation of the trajectory is obtained by eliminating the time from the two equations. We hold t = x / vo => y = h-gx²/2vo².The trajectory is a parabola because y is a second degree function of x. Bila arrives on the ground when y = 0 => t =√(2h/g)=√(2*100/9.8)=√20.4=4.51 s
The ball will fall towards the rock at a distance d = vo * t =10*4.51≈45 m
The correct answer is e