Explanation:
It is given that,
Radius of the front sprockets, 
Radius of the rear sprockets, 
The angular speed of the front sprocket is 12.3 rad/s, 
(a) Linear speed of the front sprockets, 


Linear speed of the rear sprockets, 


(b) Let
is the centripetal acceleration of the chain as it passes around the rear sprocket.




`Hence, this is the required solution.
The answer is c
Explanation:
Answer:
191.36 N/m
Explanation:
From the question,
The Potential Energy of the safe = Energy of the spring when it was compressed.
mgh = 1/2ke²............... Equation 1
Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression
Making k the subject of the equation,
k =2mgh/e²................ Equation 2
Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m
Constant: g = 9.8 m/s²
Substitute into equation 2
k = 2(1100)(9.8)(0.0024)/0.52²
k = 51.744/0.2704
k = 191.36 N/m
Hence the spring constant of the heavy-duty spring = 191.36 N/m
Answer:
sin 2θ = 1 θ=45
Explanation:
They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation
R = Vo² sin 2θ / g
Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.
We calculate the distance traveled for different angle
R = vo² Sin (2 15) /9.8
R = Vo² 0.051 m
In the table are all values in two ways
Angle (θ) distance R (x)
0 0 0
15 0.051 Vo² 0.5 Vo²/g
30 0.088 vo² 0.866 Vo²/g
45 0.102 Vo² 1 Vo²/g
60 0.088 Vo² 0.866 Vo²/g
75 0.051 vo² 0.5 Vo²/g
90 0 0
See graphic ( R Vs θ) in the attached ¡, it can be done with any program, for example EXCEL