Their linear inertia is equivalent to their masses. Let the inertia of the first moose be m₁ and the second be m₂.
m₁u + m₂u = (m₁ + m₂) x 1/3 u
3m₁ + 3m₂ = m₁ + m₂
3 m₁/m₂ + 3 = m₁/m₂ + 1
m₁/m₂ = 2
The ratio of their inertias is 2
Steam enters a cylinder—- A
A and B are equivalent. That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")
C is another way instruments are often grouped; (the "woods", the "brass")
D is another way instruments are often grouped; (the "strings", the "percussions")
Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
![B = \frac{\mu_o NI}{l} \\\\](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_o%20NI%7D%7Bl%7D%20%5C%5C%5C%5C)
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
![B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B4%5Cpi%2A10%5E%7B-7%7D%20%2A269%2A3.9%7D%7B1.02%7D%20%5C%5C%5C%5CB%20%3D%201.293%20%2A10%5E%7B-3%7D%20%5C%20T)
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T