( (77/4) + 76/2 )/2 = 28.625 km/h is what i got
Answer:
8.89 m/s² west
Explanation:
Assume east is +x. Given:
v₀ = 120 m/s
v = 0 m/s
t = 13.5 s
Find: a
v = at + v₀
0 m/s = a (13.5 s) + 120 m/s
a = -8.89 m/s²
a = 8.89 m/s² west
Answer:
Part a)
![a = 3.68 m/s^2](https://tex.z-dn.net/?f=a%20%3D%203.68%20m%2Fs%5E2)
Part b)
![a = 11.8 m/s^2](https://tex.z-dn.net/?f=a%20%3D%2011.8%20m%2Fs%5E2)
Explanation:
Part a)
For force conditions of two blocks we will have
![m_1g - T = m_1 a](https://tex.z-dn.net/?f=m_1g%20-%20T%20%3D%20m_1%20a)
![T - m_2g = m_2 a](https://tex.z-dn.net/?f=T%20-%20m_2g%20%3D%20m_2%20a)
now from above equations we have
![(m_1 - m_2) g = (m_1 + m_2) a](https://tex.z-dn.net/?f=%28m_1%20-%20m_2%29%20g%20%3D%20%28m_1%20%2B%20m_2%29%20a)
![a = \frac{m_1 - m_2}{m_1 + m_2} g](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bm_1%20-%20m_2%7D%7Bm_1%20%2B%20m_2%7D%20g)
now we know that
![m_1 = \frac{908}{9.8} = 92.65 kg](https://tex.z-dn.net/?f=m_1%20%3D%20%5Cfrac%7B908%7D%7B9.8%7D%20%3D%2092.65%20kg)
![m_2 = \frac{412}{9.8} = 42 kg](https://tex.z-dn.net/?f=m_2%20%3D%20%5Cfrac%7B412%7D%7B9.8%7D%20%3D%2042%20kg)
now from above equation we have
![a = \frac{92.65 - 42}{92.65 + 42}(9.8)](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B92.65%20-%2042%7D%7B92.65%20%2B%2042%7D%289.8%29)
![a = 3.68 m/s^2](https://tex.z-dn.net/?f=a%20%3D%203.68%20m%2Fs%5E2)
Part b)
When heavier block is removed and F = 908 N is applied at the end of the string then we have
![F - mg = ma](https://tex.z-dn.net/?f=F%20-%20mg%20%3D%20ma)
![908 - 412 = 42 a](https://tex.z-dn.net/?f=908%20-%20412%20%3D%2042%20a)
![a = 11.8 m/s^2](https://tex.z-dn.net/?f=a%20%3D%2011.8%20m%2Fs%5E2)
Answer:
1.23 m/s²
Explanation:
Given:
v₀ = 0 m/s
v = 11.1 m/s
t = 9 s
Find: a
Equation:
v = at + v₀
Plug in:
11.1 m/s = a (9 s) + 0 m/s
a = 1.23 m/s²
The runner's acceleration is 1.23 m/s².