What is the proper term for the lowest natural frequency that is produced by a vibrating object, such as an instrument?
2 answers:
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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.
Answer:
the correct answer is C, E’= 4E
Explanation:
In this exercise you are asked to calculate the electric field at a given point
E =
indicates that the field is E for r = 2m
E = (1)
the field is requested for a distance r = 1 m
E ’= k \frac{q}{r'^2}
E ’= k q / 1
from equation 1
4E = k q
we substitute
E’= 4E
so the correct answer is C