Answer
given,
before collision
mass of car A = m_a = 1300 kg
velocity of car A = v_a = 35 mph
mass of car B = m_b= 1000 kg
velocity of car B = v_b = 25 mph
after collision
V_a = 30 mph
V_b = 31.5 mph
Initial momentum
![P_1 = m_av_a + m_b v_b](https://tex.z-dn.net/?f=P_1%20%3D%20m_av_a%20%2B%20m_b%20v_b)
![P_1 = 1300 \times 35+ 1000 \times 25](https://tex.z-dn.net/?f=P_1%20%3D%201300%20%5Ctimes%2035%2B%201000%20%5Ctimes%2025)
![P_1 =70500 Kg.m/s](https://tex.z-dn.net/?f=P_1%20%3D70500%20Kg.m%2Fs)
final momentum
![P_2 = m_aV_a + m_b V_b](https://tex.z-dn.net/?f=P_2%20%3D%20m_aV_a%20%2B%20m_b%20V_b)
![P_2 = 1300 \times 30+ 1000 \times 31.5](https://tex.z-dn.net/?f=P_2%20%3D%201300%20%5Ctimes%2030%2B%201000%20%5Ctimes%2031.5)
![P_2 =70500 Kg.m/s](https://tex.z-dn.net/?f=P_2%20%3D70500%20Kg.m%2Fs)
here initial momentum is equal to the final momentum of the car.
hence, momentum is conserved in the collision.
Answer:
Physics is the science of matter and how matter interacts.
<h2>Hope it helps you...</h2>
0.8611 g/cm ^2
Density = mass/ volume
Answer:
t=0.704s
Explanation:
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?
from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed
![V^{2} =u^{2} +2as](https://tex.z-dn.net/?f=V%5E%7B2%7D%20%3Du%5E%7B2%7D%20%2B2as)
Vy^2=0+2*9.8*1.73sin40.5
Vy^2=22.021
Vy=4.69m/s
Vx^2=u^2+2*9.81*cos40.5
Vy^2=25.81
Vy=5.08m/s
V=(Vy^2+Vx^2)^0.5
V=47.71^0.5
V=6.9m/s
from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.
f=force applied
v=velocity final
u=initial velocity
m=mass of the toy, 0.046
f=ma
f=m(v-u)/t
v=u+at
6.9=0+9.8t
t=6.9/9.81
t=0.704s
Answer:
C. 1.3 x 10-11 F
I took the test and got this correct, hope this helps
Explanation: