- m1=1500kg
- m_2=3000kg
- v_1=5m/s
- v_2=7m/s
Using law of conservation of momentum
![\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5CRrightarrow%20m_1v_1-m_2v_2%3D%28m1%2Bm2%29v_3)
![\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5CRrightarrow%201500%285%29-3000%287%29%3D%281500%2B3000%29v_3)
![\\ \sf\Rrightarrow 7500-21000=4500v_3](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5CRrightarrow%207500-21000%3D4500v_3)
![\\ \sf\Rrightarrow -13500=4500v_3](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5CRrightarrow%20-13500%3D4500v_3)
![\\ \sf\Rrightarrow v_3=-3m/s](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5CRrightarrow%20v_3%3D-3m%2Fs)
72km/hr = 72000m/3600seconds(we convert km/hr to m/s
= 20m/s
Velocity= 20m/s
In 4seconds,distance covered=20x4= 80m
Therefore,an accident occurs because in 4seconds, the car would have moved past 20m
Material that is not attracted to metal
Magnitude of normal force acting on the block is 7 N
Explanation:
10N = 1.02kg
Mass of the block = m = 1.02 kg
Angle of incline Θ
= 30°
Normal force acting on the block = N
From the free body diagram,
N = mgCos Θ
N = (1.02)(9.81)Cos(30)
N = 8.66 N
Rounding off to nearest whole number,
N = 7 N
Magnitude of normal force acting on the block = 7 N
Answer:
T =176 N
Explanation:
from diagram
F -(m_1+m_2_g) = (m_1+m_2_g)a
440 - (6+4)g = (6+4)a
a =\frac{440-10*9.8}{10}
a =34.2 m/s^2
frrom free body diagram of mass m2 = 4kg
T -m_2g =m_2a
T = m_2(g +a)
T = 4(9.81+34.2)
T =176 N