Answer:
Part a)
![v_a = 3.94 m/s](https://tex.z-dn.net/?f=v_a%20%3D%203.94%20m%2Fs)
Part b)
![v_b = 3.35 m/s](https://tex.z-dn.net/?f=v_b%20%3D%203.35%20m%2Fs)
Part C)
![v_b = 6.44 m/s](https://tex.z-dn.net/?f=v_b%20%3D%206.44%20m%2Fs%20)
Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as
![a = -\frac{F_f}{m}](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7BF_f%7D%7Bm%7D)
![a = -\frac{\mu mg}{m}](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7B%5Cmu%20mg%7D%7Bm%7D)
![a = - \mu g](https://tex.z-dn.net/?f=a%20%3D%20-%20%5Cmu%20g)
now we will have
![v_f^2 - v_i^2 = 2ad](https://tex.z-dn.net/?f=v_f%5E2%20-%20v_i%5E2%20%3D%202ad)
![0 - v_i^2 = 2(-\mu g)(6.1)](https://tex.z-dn.net/?f=0%20-%20v_i%5E2%20%3D%202%28-%5Cmu%20g%29%286.1%29)
![v_a = \sqrt{(2(0.13)(9.81)(6.1)}](https://tex.z-dn.net/?f=v_a%20%3D%20%5Csqrt%7B%282%280.13%29%289.81%29%286.1%29%7D)
![v_a = 3.94 m/s](https://tex.z-dn.net/?f=v_a%20%3D%203.94%20m%2Fs)
Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have
![v_b = \sqrt{2(0.13)(9.81)(4.4)}](https://tex.z-dn.net/?f=v_b%20%3D%20%5Csqrt%7B2%280.13%29%289.81%29%284.4%29%7D)
![v_b = 3.35 m/s](https://tex.z-dn.net/?f=v_b%20%3D%203.35%20m%2Fs)
Part C)
By momentum conservation we will have
![m_1v_{1i} = m_1v_{1f} + m_2v_{2f}](https://tex.z-dn.net/?f=m_1v_%7B1i%7D%20%3D%20m_1v_%7B1f%7D%20%2B%20m_2v_%7B2f%7D)
![1400 v_b = 1100(3.94) + 1400(3.35)](https://tex.z-dn.net/?f=1400%20v_b%20%3D%201100%283.94%29%20%2B%201400%283.35%29)
![v_b = 6.44 m/s](https://tex.z-dn.net/?f=v_b%20%3D%206.44%20m%2Fs%20)
Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Answer:
speed of light simulating traveling at the speed of light. Speed of light, speed at which light waves propagate through different materials. In particular, the value for the speed of light in a vacuum is now defined as exactly 299,792,458 metres per second
The speed of sound is the distance travelled per unit of time by a sound wave as it propagates through an elastic medium
Answer:
<em>A</em><em>.</em><em>increases</em><em> </em><em>friction</em>
Explanation:
3. is the answer, <span>Sodium needs to lose one electron, and chlorine needs to gain one electron. This is because Sodium's row always wants to give away an electron, while Chlorine's row wants to gain an electron.</span>