The speed of the toy when it hits the ground is 2.97 m/s.
The given parameters;
- mass of the toy, m = 0.1 kg
- the maximum height reached by the, h = 0.45 m
The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.
P.E = K.E
![mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}](https://tex.z-dn.net/?f=mgh_%7Bmax%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv_%7Bmax%7D%5E2%5C%5C%5C%5Cgh_%7Bmax%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20v_%7Bmax%7D%5E2%5C%5C%5C%5Cv_%7Bmax%7D%5E2%3D%202gh_%7Bmax%7D%5C%5C%5C%5Cv_%7Bmax%7D%20%3D%20%5Csqrt%7B2gh_%7Bmax%7D%7D)
Substitute the given values and solve the speed;
![v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20%5Csqrt%7B2%5Ctimes%209.8%20%5Ctimes%200.45%7D%20%5C%5C%5C%5Cv_%7Bmax%7D%20%3D%202.97%20%5C%20m%2Fs)
Thus, the speed of the toy when it hits the ground is 2.97 m/s.
Learn more here: brainly.com/question/7562874
During the collision between two balls on the pool table there is no external force along the line of collision between them
Since there is no external force on it so here we can say
![F = 0 = \frac{\Delta P}{\Delta t}](https://tex.z-dn.net/?f=F%20%3D%200%20%3D%20%5Cfrac%7B%5CDelta%20P%7D%7B%5CDelta%20t%7D)
here we have
![\Delta P = 0](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%200)
so we can say
![P_i = P_f](https://tex.z-dn.net/?f=P_i%20%3D%20P_f)
since there is no external force so we can say during the collision the momentum of two balls will remain conserved
Well,
When an object's velocity changes, we call it acceleration.
Acceleration: The time rate of change in an object's velocity
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