Answer:
its not moving at a constant velocity because it is slowing down
Explanation:
<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>
If time is the x axis and distance is the y axis then yes, in the case that time is going by but distance remains the same.
Answer:
m = 1,975 m / kg
, b = 38.05 m
Explanation:
In this experiment, the elongation is plotted against the applied mass
getting a straight line
y = m x + b
where b would be the initial length of spring let's calculate the slope for which we use two well separated points
m = (56.3 -48.4) / (8 - 4)
m = 1,975 m / kg
the equation remains
y = 1,975 x + b
for x = 2 kg y = 42.0 m
we substitute in the equation
42 = 1,975 2 + b
b = 42 - 3.95
b = 38.05 m