air beomes more dense
below dewpoint = a dew forms ... wet underfoot.
cold enough = frost ???
Explanation:
It is given that,
The velocity of a particle is given by :

Where
v is in m/s and t is in seconds
Let a is the acceleration of the object at time t. So,



When a = 0

t = 2.5 s
a is zero at t = 2.5 s. Velocity, 
v = -75 m/s
Since,
, s is the distance travelled



At t = 2.5 s, 
s = −83.34 m
Hence, this is the required solution.
Answer:
It's 1.0000042 times longer in summer than in winter. It represents a 1.6 centimeters difference between seasons.
Explanation:
The linear coefficient of thermal expansion for steel is about
. From the equation of linear thermal expansion, we have:

Taking the winter day as the initial, and the summer day as the final, we can take the relationship between them:
![L_{summer}=L_{winter}[1+(1.2*10^{-7}\°C^{-1})(30\°C+5\°C)]\\\\L_{summer}=(1.0000042)L_{winter}](https://tex.z-dn.net/?f=L_%7Bsummer%7D%3DL_%7Bwinter%7D%5B1%2B%281.2%2A10%5E%7B-7%7D%5C%C2%B0C%5E%7B-1%7D%29%2830%5C%C2%B0C%2B5%5C%C2%B0C%29%5D%5C%5C%5C%5CL_%7Bsummer%7D%3D%281.0000042%29L_%7Bwinter%7D)
It means that the bridge is 1.0000042 times longer in summer than in winter. If we multiply it by the length of the bridge, we obtain that the difference is of about 1.6 centimeters between the two seasons.
Answer:
By decreasing the surface area exposed to the air. Vertical body position instead of horizontal.
Explanation:
So the terminal velocity formula is as follows.
v = sqrt[(2*m*g)/(p*A*c)]
m = mass
g = 9.81 m/s^2
p = density of air
A = surface area of object
c = Drag coefficient
So the only thing you can change in mid-air is surface area.