Answer: 2.1 × 10^7 m/s
Explanation:
Please see the attachments below
Answer:
The average power the woman exerts is 0.5 kW
Explanation:
We note that power, P = The rate at which work is done = Work/Time
Work = Energy
The total work done is the potential energy gained which is the energy due to vertical displacement
Given that the vertical displacement = 5.0 m, we have
Total work done = Potential energy gained = Mass, m × Acceleration due to gravity, g × Vertical height, h
m = 51 kg
g = Constant = 9.81 m/s²
h = 5.0 m
Also, time, t = 5.0 s
Total work done = 51 kg × 9.81 m/s²× 5 m = 2501.55 kg·m²/s² = 2501.55 J
P = 2501.55 J/(5 s) = 500.31 J/s = 500.31 W ≈ 500 W = 0.5 kW.
Answer:
v = 14.32 m/s
Explanation:
According to the principle of conservation of linear momentum, both the momentum and kinetic energy of the system are conserved. Since the two balls are in the same direction of motion before collision, then;
+
= (
+
) v
0.035 × 12 + 0.120 × 15 = (0.035 + 0.120) v
0.420 + 1.800 = (0.155) v
2.22 = 0.155 v
⇒ v = ![\frac{2.22}{0.155}](https://tex.z-dn.net/?f=%5Cfrac%7B2.22%7D%7B0.155%7D)
= 14.323
The velocity of the balls after collision is 14.32 m/s.
Answer:
14.8 kg
Explanation:
We are given that
![m_1=43.7 kg](https://tex.z-dn.net/?f=m_1%3D43.7%20kg)
![m_2=12.1 kg](https://tex.z-dn.net/?f=m_2%3D12.1%20kg)
![g=9.8 m/s^2](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
![a=\frac{1}{2}(9.8)=4.9 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B1%7D%7B2%7D%289.8%29%3D4.9%20m%2Fs%5E2)
We have to find the mass of the pulley.
According to question
![T_2-m_2 g=m_2 a](https://tex.z-dn.net/?f=T_2-m_2%20g%3Dm_2%20a)
![T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N](https://tex.z-dn.net/?f=T_2%3Dm_2a%2Bm_2g%3Dm_2%28a%2Bg%29%3D12.1%289.8%2B4.9%29%3D177.87%20N)
![T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N](https://tex.z-dn.net/?f=T_1%3Dm_1%28g-a%29%3D43.7%289.8-4.9%29%3D214.13%20N)
Moment of inertia of pulley=![I=\frac{1}{2}Mr^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7B2%7DMr%5E2)
![(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)](https://tex.z-dn.net/?f=%28T_2-T_1%29r%3DI%28-%5Calpha%29%3D%5Cfrac%7B1%7D%7B2%7DMr%5E2%28%5Cfrac%7B-a%7D%7Br%7D%29%3D%5Cfrac%7B1%7D%7B2%7DMr%28-4.9%29)
Where ![\alpha=\frac{a}{r}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7Ba%7D%7Br%7D)
![(177.87-214.13)=-\frac{1}{2}(4.9)M](https://tex.z-dn.net/?f=%28177.87-214.13%29%3D-%5Cfrac%7B1%7D%7B2%7D%284.9%29M)
![-36.26=-\frac{1}{2}(4.9)M](https://tex.z-dn.net/?f=-36.26%3D-%5Cfrac%7B1%7D%7B2%7D%284.9%29M)
![M=\frac{36.26\times 2}{4.9}=14.8 kg](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B36.26%5Ctimes%202%7D%7B4.9%7D%3D14.8%20kg)
Hence, the mass of the pulley=14.8 kg
The correct answer would be yes