Answer:
T=575.16K
Explanation:
To solve the problem we proceed to use the 1 law of diffusion of flow,
Here,

is the rate in concentration
is the rate in thickness
D is the diffusion coefficient, where,

Replacing D in the first law,

clearing T,

Replacing our values



Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.
Answer: well the time it takes to fall 100m is the same time it takes to travel 65m horizontally.
The time to fall vertically, t is sqrt(2d/g). [this comes from d = 1/2at^2]
so t = sqrt(2*100/9.8) = 4.52s.
The vertical speed at that time is g*t = 9.8*4.52 = 44.3m/s
The horizontal speed is the horizontal distance over the same 4.52s, = 65/4.52 = 14.4m/s.
so the final velocity is = sqrt(44.3^2 + 14.4^2) = 46.6m/s
Explanation: yes