Answer:
R = 0.0503 m
Explanation:
This is a projectile launching exercise, to find the range we can use the equation
R = v₀² sin 2θ / g
How we know the maximum height
² =
² - 2 g y
= 0
= √ 2 g y
= √ 2 9.8 / 15
= 1.14 m / s
Let's use trigonometry to find the speed
sin θ = / vo
vo = / sin θ
vo = 1.14 / sin 60
vo = 1.32 m / s
We calculate the range with the first equation
R = 1.32² sin(2 60) / 30
R = 0.0503 m
When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.
During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.
Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.
Answer:
The value is 
Explanation:
From the question we are told that
The angular speed is 
The distance between the minimum and maximum external position is 
Generally the amplitude of the crank shaft is mathematically represented as
=> 
=> 
Generally the maximum speed of the piston is mathematically represented as
=> 
=>
