Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
![v_{f}^{2}=v_{i}^{2} +2*a*x](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3Dv_%7Bi%7D%5E%7B2%7D%20%2B2%2Aa%2Ax)
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)
Answer:
Answer is in the following attachment.
Explanation:
The mass contributes with the time of thermal energy transfer with respect to the material type but most importantly the material type will determine rate at which the material absorbs the transfer of heat or thermal energy by either three types, conduction, convection and radiation.
1) 3 miles/Hour
The speed is defined as the distance covered divided by the time taken:
![v=\frac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D)
where
d = 1.5 mi is the distance
t = 0.5 h is the time taken
Substituting,
![v=\frac{1.5}{0.5}=3 mi/h](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B1.5%7D%7B0.5%7D%3D3%20mi%2Fh)
2) 1.34 m/s south
Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:
is the displacement in meters
is the time taken in seconds
Substituting,
![v=\frac{2414 m}{1800 s}=1.34 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2414%20m%7D%7B1800%20s%7D%3D1.34%20m%2Fs)
And the direction of the velocity is the same as the displacement, so it is south.