The tension in the string at the 60-cm mark will be 0.9 N.The tension force is applied along the whole length of the wire, pulling energy equally on both ends.
<h3>What is tension force?</h3>
The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.
The given data in the problem is;
The length of the stick is, L= 1.20 m =120 cm
T is the tension in the string at the 60-cm mark;
T₁ is the tension at the 20-cm mark
From the torque equilibrium law;
Hence the tension in the string at the 60-cm mark will be 0.9 N.
To learn more about the tension force refer to the link;
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Because of conservation of energy, ep is equal to ek.
at that point ep=ek
the kinetic energy would be 32,781J
Answer:
V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s
S = 1/2 g t^2 since initial speed is zero
S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m
Answer: acceleration = 3.27m/s^2
Explanation:
Given that the
Mass M = 44kg
Angle Ø = 15 degree
Coefficient of friction ų = 0.7
Force F = 222N
F - Fr = ma ...... (1)
Where Fr = frictional force
Fr = ųN
N = normal reaction = mg
Fr = ųmgsinØ
Fr = 0.7 × 44 × 9.81 × sin 15
Fr = 78.2N
Substitutes Fr, F and M into equation one.
222 - 78.2 = 44a
143.79 = 44a
Make a the subject of formula
a = 143.79/44
Acceleration a = 3.27 m/s^2
