Answer:
(1) Support
(2) Tension
(3) Reaction
The constraint forces are forces that occur in pairs such that the net force (effect) is zero
The details of the constraint forces are;
(1) The support force at the top of the swing acting upwards
(2) The tension in the swing rope acting upwards
(3) The (normal) reaction between the legs of the swing frame and the ground
Explanation:
The normal reaction is the force equal in magnitude and opposite in direction to the total weight force the swing and the person on the swing posses due to gravitational attraction which prevents the downward or sideways motion of the entire swing
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
Answer:
displacement at 45 s = 30
65 s = 50
So the average speed over the interval from 45 s to 65 s is
(50 - 30) cm / 20 s = 1 cm / sec
As a check an average speed of 1 cm / sec for 20 sec will produce a
displacement of 1 cm / sec * 20 sec = 20 cm or from 30 to 50 cm
Answer:
1.25 m
Explanation:
From the question given above, the following data were obtained:
Force ratio = 2.5
Distance of load from the fulcrum = 0.5 m
Distance of effort =.?
The distance of the effort from the fulcrum can be obtained as illustrated below:
Force ratio = Distance of effort / Distance of load
2.5 = Distance of effort / 0.5
Cross multiply
Distance of effort = 2.5 × 0.5
Distance of effort = 1.25 m
Therefore, the distance of the effort from the fulcrum is 1.25 m