<span>First let's find the acceleration required in the barrel to speed the ball up from 0 to 83 m/s in a distance of 2.17 m. We know the force the cannon exerts on the cannonball is 20000 N; if we can find this acceleration then we can use F = ma to find the mass.
We can find the acceleration using one of the kinematic equations of motion. We have:
u = initial speed = 0 m/s
v = final speed = v0 = 83 m/s
d = distance = 2.17 m
a = acceleration = ?
v² = u² + 2ad. Since u = 0, this reduces to v² = 2ad and rearranges to a = v²/2d = 83²/2*2.17 = 83²/4.34 = 1587.327 m/s².
Now F = ma, so m = F/a = (20000N)/(1587.327 m/s²) = 12.6 kg.
For part 2, use the Range Equation:
If R is the horizontal distance the cannonball travels,
v = v0 = the initial velocity = 83 m/s
g = acceleration due to gravity - 9.8 m/s²
x the launch angle relative to the horizontal, then
R = (v²sin(2x))/g.
So R = (83²sin(2*37))/9.8
= (6889sin74)/9.8 = 676 m.
So the target ship is 676 m away.</span>
Energy is transferred.
<h3>What is Work?</h3>
Every time labour is performed, energy is transmitted, as shown by the definition of work in physics. Work is the result of the displacement's magnitude and the component of force acting in that direction.
What are 3 types of work done?
- Positive
- Negative
- Zero work done.
Work is the result of force and distance, or the force exerted over a specific distance. In other words, Work = Force x Distance.
Joules are used to measure work.
1. Positive Work: Positive work is when a force moves an object in that direction. The motion of a ball descending toward the earth while it is displaced in the direction of the force of gravity is an illustration of this type of labour.
2. Negative Work: The work is said to be negative if the force and the displacement are in the opposing directions. For instance, if a ball is thrown high, its displacement will also be upwards, while the force from the earth's gravity will be downward.
3. Zero Work: When the force and displacement directions are parallel to one another, there is no work done on the item by the force.
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Answer:
v = 27.3 m/s
Explanation:
Given that
Acceleration ,a= 4.2 m/s²
Time ,t= 6.5 s
Lets take the maximum speed gain by Thomson's= v
We know that ,if acceleration is constant then the speed v is given as
v= u + a t
v=final speed
u=initial speed
a=acceleration
t=time
Here the initial speed of Thomson's ,u = 0 m/s
Now by putting the values in the above equation we get
v= 0 + 4.2 x 6.5 m/s
v = 27.3 m/s
Therefore the maximum speed gain by Thomson will be 27.3 m/s.