Answer:
The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A
Explanation:
Force on a wire carrying current in an electric field is given by
F = (B)(I)(L) sin θ
For this question,
The magnetic force must match the weight of the wire.
F = mg
mg = (B)(I)(L) sin θ
(m/L)g = (B)(I) sin θ
Mass per unit length = 75 g/m = 0.075 kg/m
B = magnetic field = 0.12 T
I = ?
g = acceleration due to gravity = 9.8 m/s
θ = angle between wire's current direction and magnetic field = 90°
0.075 × 9.8 = 0.12 × I sin 90°
I = 0.075 × 9.8/0.12 = 6.125 A
Answer:
deductive reasoning usually follows steps .
- That is, how we predict what the observations should be if the theory were correct
Answer:
The answer is the 1st one
To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as
![f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
Here,
k = Spring constant
m = Mass
Our values are given as,
![m = 35g = 35*10^{-3}kg](https://tex.z-dn.net/?f=m%20%3D%2035g%20%3D%2035%2A10%5E%7B-3%7Dkg)
![f = 1 Hz](https://tex.z-dn.net/?f=f%20%3D%201%20Hz)
Rearranging to find the spring constant we have that,
![k = (2\pi f \sqrt{m})^2](https://tex.z-dn.net/?f=k%20%3D%20%282%5Cpi%20f%20%5Csqrt%7Bm%7D%29%5E2)
![k = 4\pi^2 f^2 m](https://tex.z-dn.net/?f=k%20%3D%204%5Cpi%5E2%20f%5E2%20m)
![k = (4) (\pi)^2 (1) (35*10^{-3})](https://tex.z-dn.net/?f=k%20%3D%20%284%29%20%28%5Cpi%29%5E2%20%281%29%20%2835%2A10%5E%7B-3%7D%29)
![k = 1.38N/m](https://tex.z-dn.net/?f=k%20%3D%201.38N%2Fm)
Therefore the spring constant is 1.38N/m