A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?
1 answer:
Answer:
v = 70 m/s
Explanation:
Range on the cannon ball is given as
here the angle of the projection of the ball is given as 45 degree
now we know that if the velocity of the ball is "v" then its two components will be given as
so here time of flight of the motion is given as
also the range is given as
now plug in all data in this equation
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It depends on both of them.
In fact, the projectile begins its motion with an initial velocity of
and an angle of 
. On the y-axis (vertical direction), it is an accelerated motion with acceleration equal to -g (gravitational acceleration). The vertical velocity of the projectile at any time t is given by
and as it can be seen, this depends on both initial velocity and angle.
In fact, the projectile begins its motion with an initial velocity of
and as it can be seen, this depends on both initial velocity and angle.
Answer:
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Explanation:
The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is
K= constant depends on the system of units used.
W= weight =485 g
D= density =7.9 g/cm³
A = exposed specimen area =100 in² =6.452 cm²
K=534 to give CPR in mpy
K=87.6 to give CPR in mm/yr
mpy
=37.4mpy
mm/yr
=0.952 mm/yr
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.