Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s
v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m
Answer:
¨Facts you didn´t know¨ or ¨unknown facts¨
Explanation:
Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is
The temperature of water is
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)


Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.

where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)