Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
Because AC emf is a sine wave
Decreases. Air resistance will slow a falling object to its terminal velocity, placing a limit on its acceleration.
Answer:
The angle is 
Explanation:
From the question we are told that
The distance of the dartboard from the dart is 
The time taken is 
The horizontal component of the speed of the dart is mathematically represented as
where u is the the velocity at dart is lunched
so
substituting values
=> 
From projectile kinematics the time taken by the dart can be mathematically represented as
=> 
=> 
![\theta = tan^{-1} [0.277]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%20%5B0.277%5D)
Answer:
18.63 N
Explanation:
Assuming that the sum of torques are equal
Στ = Iα
First wheel
Στ = 5 * 0.51 = 3 * (0.51)² * α
On making α subject of formula, we have
α = 2.55 / 0.7803
α = 3.27
If we make the α of each one equal to each other so that
5 / (3 * 0.51) = F2 / (3 * 1.9)
solve for F2 by making F2 the subject of the formula, we have
F2 = (3 * 1.9 * 5) / (3 * 0.51)
F2 = 28.5 / 1.53
F2 = 18.63 N
Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.
