A: The total building of Campbell high school, including the trailers and the construction area
Answer:
avriage force F = 2722.5 N
Explanation:
For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.
vf² = v₀² - 2 ax
The final carriage speed is zero (vf = 0)
0 = v₀² - 2ax
a = v₀² / 2x
a = 1.1²/(2 0.200)
a = 3.025 m / s²
a = 3.0 m/s²
We calculate the average force
F = ma
F = 900 3,025
F = 2722.5 N
Answer:
1000 kgm²/s, 400 J
1000 kgm²/s, 1000 J
600 J
Explanation:
m = Mass of astronauts = 100 kg
d = Diameter
r = Radius = ![\frac{d}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B2%7D)
v = Velocity of astronauts = 2 m/s
Angular momentum of the system is given by
![L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s](https://tex.z-dn.net/?f=L%3Dmvr%2Bmvr%5C%5C%5CRightarrow%20L%3D2mvr%5C%5C%5CRightarrow%20L%3D2%5Ctimes%20100%5Ctimes%202%5Ctimes%202.5%5C%5C%5CRightarrow%20L%3D1000%5C%20kgm%5E2%2Fs)
The angular momentum of the system is 1000 kgm²/s
Rotational energy is given by
![K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J](https://tex.z-dn.net/?f=K%3DI%5Comega%5E2%5C%5C%5CRightarrow%20K%3D%5Cfrac%7B1%7D%7B2%7D%28mr%5E2%29%5Cleft%28%5Cfrac%7Bv%7D%7Br%7D%5Cright%29%5E2%5C%5C%5CRightarrow%20K%3Dmv%5E2%5C%5C%5CRightarrow%20K%3D100%5Ctimes%202%5E2%5C%5C%5CRightarrow%20K%3D400%5C%20J)
The rotational energy of the system is 400 J
There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s
![L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s](https://tex.z-dn.net/?f=L_i%3DL_f%5C%5C%5CRightarrow%202mv_ir_i%3D2mv_fr_f%5C%5C%5CRightarrow%20v_f%3D%5Cfrac%7Bv_ir_i%7D%7Br_f%7D%5C%5C%5CRightarrow%20v_f%3D%5Cfrac%7B2%5Ctimes%202.5%7D%7B0.5%7D%5C%5C%5CRightarrow%20v_f%3D10%5C%20m%2Fs)
Energy
![E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J](https://tex.z-dn.net/?f=E_2%3Dmv_f%5E2%5C%5C%5CRightarrow%20E_2%3D100%5Ctimes%2010%5C%5C%5CRightarrow%20E_2%3D1000%5C%20J)
The new energy will be 1000 J
Work done will be the change in the kinetic energy
![W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J](https://tex.z-dn.net/?f=W%3DE_2-E%5C%5C%5CRightarrow%20W%3D1000-400%5C%5C%5CRightarrow%20W%3D600%5C%20J)
The work done is 600 J
27.5 because of you divide the 55miles with the time you get your velocity which is the speed.