Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
Answer:
Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz
Explanation:
Given data
Source Frequency fs=600Hz
Length r=1.0m
RPM=100 rpm
The speed of the generator is calculated as:

Substitute the given values

For approaching generator the frequency is calculated as:

On the other hand,for the receding generator,Doppler's effect is expressed as:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz
Answer:
Explanation:
From the question we are told that
The moment of inertia is 
The final angular speed is 
The time taken is 
The initial angular speed is 
Generally the average angular acceleration is mathematically represented as

=> 
=> 
Generally the torque is mathematically represented as

=> 
=> 
Answer:
208 Joules
Explanation:
The radius of the circular path the charge moves, r = 26 m
The magnetic force acting on the charge particle, F = 16 N
Centripetal force,
= m·v²/r
Kinetic energy, K.E. = (1/2)·m·v²
Where;
m = The mass of the charged particle
v = The velocity of the charged particle
r = The radius of the path of the charged particle
Whereby the magnetic force acting on the charge particle = The centripetal force, we have;
F =
= m·v²/r = 16 N
(1/2) × r ×
= (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.
∴ (1/2) × r ×
= (1/2) × 26 m × 16 N = = (1/2)·m·v² = K.E.
∴ 208 Joules = K.E.
The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.