Answer:
(a) the runner's kinetic energy at the given instant is 308 J
(b) the kinetic energy increased by a factor of 4.
Explanation:
Given;
mass of the runner, m = 64.1 kg
speed of the runner, u = 3.10 m/s
(a) the kinetic energy of the runner at this instant is calculated as;
![K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J](https://tex.z-dn.net/?f=K.E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mu%5E2%5C%5C%5C%5CK.E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%2064.1%20%5Ctimes%203.1%5E2%5C%5C%5C%5CK.E_i%20%3D%20308%20%5C%20J)
(b) when the runner doubles his speed, his final kinetic energy is calculated as;
![K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J](https://tex.z-dn.net/?f=K.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mu_f%5E2%5C%5C%5C%5CK.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%282u%29%5E2%5C%5C%5C%5CK.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2064.1%20%5C%20%5Ctimes%20%282%5Ctimes%203.1%29%5E2%5C%5C%5C%5CK.E_f%20%3D%201232%20%5C%20J)
the change in the kinetic energy is calculated as;
![\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4](https://tex.z-dn.net/?f=%5Cfrac%7BK.E_f%7D%7BK.E_i%7D%20%3D%20%5Cfrac%7B1232%7D%7B308%7D%20%3D4)
Thus, the kinetic energy increased by a factor of 4.
Answer:
Centre of mass of any body is a point where all mass of a body is supposed to be concentrated
it lies in geometrical centre....
<span>Homicide detectives are hired to find murderers.
hope it helps.</span>
Answer:
a) t = 4.04 s
, b) x = 89.77 m
Explanation:
This is a one-dimensional kinematics exercise.
a) Let's use caution to find the time, the vehicle speed when stopped is zero
v = v₀ - at
let's reduce the magnitudes to the SI system
v₀ = 160 km /h (1000m / 1km) (1h / 3600s) = 44.44 m / s
0 = v₀ - at
t = v₀ / a
t = 44.44 / 11
t = 4.04 s
b) to calculate the distance traveled we use
x = v₀t - ½ a t²
x = 44.44 4.04 - ½ 11.0 4.04²
x = 179.54 - 89.77
x = 89.77 m