Answer:
2991.47 [cm^2]
Explanation:
To solve this problem we must perform a dimensional analysis and use the corresponding conversion values:
![3.22[ft^{2}]*\frac{12^{2}in^{2} }{1^{2}ft^{2}} *\frac{2.54^{2}cm^{2} }{1^{2}in^{2} } \\2991.47[cm^{2}]](https://tex.z-dn.net/?f=3.22%5Bft%5E%7B2%7D%5D%2A%5Cfrac%7B12%5E%7B2%7Din%5E%7B2%7D%20%7D%7B1%5E%7B2%7Dft%5E%7B2%7D%7D%20%2A%5Cfrac%7B2.54%5E%7B2%7Dcm%5E%7B2%7D%20%20%7D%7B1%5E%7B2%7Din%5E%7B2%7D%20%7D%20%5C%5C2991.47%5Bcm%5E%7B2%7D%5D)
It creates friction on the forward moving object, causing it to loose momentum, until finally, it stops.
Hope this helps!
99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
= (0²-27.5²)/(2x50.0)
=-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!
Answer:
v = 2.029 m/s
Explanation:
Given
L = 84.0 cm ⇒ R = 0.5*L = 0.5*84 cm = 42 cm = 0.42 m
m₁ = 0.600 kg
m₂ = 0.200 kg
g = 9.8 m/s²
u₁ = u₂ = 0 m/s
v₁ = ?
v₂ = ?
Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).
In this system mechanical energy is conserved, so we can match its value in the horizontal position with the one in the vertical.
then
Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J
Ef = Kf + Uf
⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²
⇒ Uf = m₁*g*h₁ + m₂*g*h₂ = 0.6*9.8*(-0.42) + 0.2*9.8*0.42 = - 1.6464
⇒ Ef = Kf + Uf = 0.4*v² - 1.6464
Since
0 = 0.4*v² - 1.6464 ⇒ v = 2.029 m/s
v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.
v₁ = v₂ = v ⇒ ω₁*R₁ = ω₂*R₂ ⇒ ω₁*R = ω₂*R ⇒ ω₁ = ω₂ = ω
⇒ v = ω*R
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv
