Acceleration
On a velocity time graph, the y-axis is velocity, and the x-axis is time. So the slope is the change in velocity for every unit of change in time.
acceleration- the rate of change of velocity per unit of time.
hope this helps :)
Explanation:
To my understanding of comic books, weak force indicates that the hero was strong enough to overcome its powers...
Assuming both billiard balls have the same mass, conservation of momentum says
![m\vec v_1 + m\vec v_2 = m{\vec v_1}\,' + m{\vec v_2}\,'](https://tex.z-dn.net/?f=m%5Cvec%20v_1%20%2B%20m%5Cvec%20v_2%20%3D%20m%7B%5Cvec%20v_1%7D%5C%2C%27%20%2B%20m%7B%5Cvec%20v_2%7D%5C%2C%27)
where m = mass of both billiard balls, and v₁ and v₂ = their initial velocities, and v₁' and v₂' = their final velocities. The masses are the same so the exact value of m is irrelevant. The first ball has initial speed 5 m/s and the second is at rest, so
![\left(5 \dfrac{\rm m}{\rm s}\right) \, \vec\imath = {\vec v_1}\,' + {\vec v_2}\,'](https://tex.z-dn.net/?f=%5Cleft%285%20%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%20%5C%2C%20%5Cvec%5Cimath%20%3D%20%7B%5Cvec%20v_1%7D%5C%2C%27%20%2B%20%7B%5Cvec%20v_2%7D%5C%2C%27)
After the collision, the first ball has speed 4.35 m/s and is moving at angle of 30° below the original path, so
![{\vec v_1}\,' = \left(4.35\dfrac{\rm m}{\rm s}\right)\left(\cos(30^\circ) \, \vec\imath + \sin(30^\circ) \, \vec\jmath\right) \approx \left(3.77 \dfrac{\rm m}{\rm s}\right) \vec\imath + \left(-2.18 \dfrac{\rm m}{\rm s}\right) \vec\jmath](https://tex.z-dn.net/?f=%7B%5Cvec%20v_1%7D%5C%2C%27%20%3D%20%5Cleft%284.35%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%5Cleft%28%5Ccos%2830%5E%5Ccirc%29%20%5C%2C%20%5Cvec%5Cimath%20%2B%20%5Csin%2830%5E%5Ccirc%29%20%5C%2C%20%5Cvec%5Cjmath%5Cright%29%20%5Capprox%20%5Cleft%283.77%20%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%20%5Cvec%5Cimath%20%2B%20%5Cleft%28-2.18%20%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%20%5Cvec%5Cjmath)
Then the second ball has final velocity vector
![{\vec v_2}\,' \approx \left(1.23 \dfrac{\rm m}{\rm s}\right) \vec\imath + \left(2.18 \dfrac{\rm m}{\rm s}\right) \vec\jmath](https://tex.z-dn.net/?f=%7B%5Cvec%20v_2%7D%5C%2C%27%20%5Capprox%20%5Cleft%281.23%20%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%20%5Cvec%5Cimath%20%2B%20%5Cleft%282.18%20%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%20%5Cvec%5Cjmath)
so it moves with speed
![\left\|{\vec v_2}\,'\right\| \approx \sqrt{\left(1.23\dfrac{\rm m}{\rm s}\right)^2 + \left(2.18\dfrac{\rm m}{\rm s}\right)^2} \approx \boxed{2.50 \dfrac{\rm m}{\rm s}}](https://tex.z-dn.net/?f=%5Cleft%5C%7C%7B%5Cvec%20v_2%7D%5C%2C%27%5Cright%5C%7C%20%5Capprox%20%5Csqrt%7B%5Cleft%281.23%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%5E2%20%2B%20%5Cleft%282.18%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%5E2%7D%20%5Capprox%20%5Cboxed%7B2.50%20%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%7D)
at an angle of
![\theta \approx \tan^{-1}\left(\dfrac{2.18}{1.23}\right) \approx \boxed{60.5^\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%5Capprox%20%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac%7B2.18%7D%7B1.23%7D%5Cright%29%20%5Capprox%20%5Cboxed%7B60.5%5E%5Ccirc%7D)
or about 60.5° above the original line of motion.
Answer:
36 N
Explanation:
The tension on the string :
T =mv²/r ------where
m= mass of object = 8 kg
v= tangential velocity = 3 m/s
r= length of the string = 2m
T= [ 8 * 3² ] / 2
T= 36 N