<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback
So for conservation of momentum,
rho = mv
M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf
For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2
Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
Acceleration of the car is given by
Mass is given as m = 2580 kg
Now force is given by
Now if air bags are open and the time is increased to t = 1 minute.
acceleration is given by
Now force is given by
So the force is reduced when air bags are opened.
There are several formulas that describe the distance, speed, and time for
a falling body. The one I use the most happens to be the one that'll be the
most useful to solve this problem:
Distance = 1/2 g t²
We know the distance and we know ' g ', so we can use
this formula to find ' t '.
Distance = (1/2) (gravity) (time)²
(239 m) = (1/2) (3.7 m/s²) (time)²
Divide each side
by 1.85 m/s² : (129 m) / (1.85 m/s²) = (time)²
(129/1.85) sec² = (time)²
Take the square root
of each side: 8.35 sec = time
Answer:
k=19 N/m
Explanation:
Use the equation F=kx for this problem and isolate k since that is what you're solving for (k=F/x). Plug in your values (F=1.2 N and x=0.064 m) and solve for k which is 19 N/m.