You should calculate 40 kg and the radius 3mm.
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:
The tension to bring the guitar string into tune is 372.95 Hz.
Explanation:
Given;
current frequency, f₁ = 248 Hz
current tension, T₁ = 350 N
fundamental frequency, f₂ = 256
The tension on the string to bring the guitar string into tune is calculated as;
Therefore, the tension to bring the guitar string into tune is 372.95 Hz.
Answer:
huh? do you need help on math?
Explanation:
what do you mean?
Answer:
(d) III only
Explanation:
We have to observe the motion of the bag with respect to taxi , considering taxi as stationary or inertial frame . Since bag is not moving with respect to taxi , the inertial frame that means , net force on it is zero .So option i and ii are ruled out .
Now how to explain motion of the bag ie why it is stationary ie what are the balancing force acting on it. We know that on a body on circular path , a force called centripetal force is acting on it . So that force must be acting on it . The balancing force is the frictional force which is keeping it stationary with respect to taxi . Hence the third option is correct.
