Can someone please answer this pleaseee!!!!!!!!
1 answer:
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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
Answer:
Tension, T = 0.0115 N
Explanation:
Given that,
Mass of the plastic ball, m = 1.1 g
Length of the string, l = 56 cm
A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :
T is tension in the string
So, the tension in the string is 0.0115 N.
