What scale can I use for two force 0.8 Newtons and 0.9 Newtons? Someone please help
1 answer:
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centripetal acceleration is given by formula
given that
now we have
so the ratationa frequency is given by
Refer to the diagram shown below.
R = total resistive force.
80% of this force is experienced by the boat and trailer.
T = the force in the hitch between the car and the trailer.
Overall motion:
Total mass = 1100 + 700 = 1800 kg
Acceleration = 0.55 m/s²
Driving force = 1900 N
Resisting force = R N
Therefore
(1900 - R N) = (1800 kg)*(0.55 m/s²)
1900 - R = 990
R = 910 N
0.2R = 182 N
0.8R = 728 N
Consider the motion of the car:
Driving force = 1900 N
Resistive force = 0.2R = 182 N
Force in the hitch = T NAcceleration = 0.55 m/s²
Mass = 1100 kg
Therefore
1900 - 0.2R - T = 1100*0.55
1900 - 182 -T = 605
T = 1113 N
Answer:
R = 910 N, the total resistive force.
T = 1113 N, the force in the hitch between the car and trailer.
R = total resistive force.
80% of this force is experienced by the boat and trailer.
T = the force in the hitch between the car and the trailer.
Overall motion:
Total mass = 1100 + 700 = 1800 kg
Acceleration = 0.55 m/s²
Driving force = 1900 N
Resisting force = R N
Therefore
(1900 - R N) = (1800 kg)*(0.55 m/s²)
1900 - R = 990
R = 910 N
0.2R = 182 N
0.8R = 728 N
Consider the motion of the car:
Driving force = 1900 N
Resistive force = 0.2R = 182 N
Force in the hitch = T NAcceleration = 0.55 m/s²
Mass = 1100 kg
Therefore
1900 - 0.2R - T = 1100*0.55
1900 - 182 -T = 605
T = 1113 N
Answer:
R = 910 N, the total resistive force.
T = 1113 N, the force in the hitch between the car and trailer.