<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>
Answer:
W = 46 J
Explanation:
We need to find the angle between the two vectors Force vector and displacement vector.
First we will find the angle α of the force vector
Then we find the angle β of the displacement vector
With these two angles we can find the angle between the two vectors
∅ = α + β = 25.56 deg
The definition of work is given by the expression
The absolute value of F will be:
The absolute value of d will be:
Now we have:
Answer:
0.22 b
Explanation:
Quadrupole moment of the nucleon is,
And also,
And,
Now,
For Bismuth and A is 209.
Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b
Answer:
<h2>To find the resultant force <u>subtract the magnitude of the smaller force from the magnitude of the larger force.</u> The direction of the resultant force is in the same direction as the larger force. A force of 5 N acts to the right, and a force of 3 N act to the left. Calculate the resultant force.</h2>
conducted
Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more. These molecules then bump into nearby particles and transfer some of their energy to them. This then continues and passes the energy from the hot end down to the colder end of the substance.