Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,
Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.
v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s
Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2
Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds
Then,
x = 8*0.553 = 4.424 m
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Answer:
medium
Explanation:
<em>A sound </em><em>medium</em><em> is defined as channel through which sound can travel or be transmitted. </em>
Sound medium could be in the form gases, liquids, solids or plasmas. Space is made up of vacuum and therefore, has no medium within it. Hence, space cannot transmit sound in any form or allows sound to travel through it.
Assuming the accleration applied was constant, we have
Then the force applied to the ball is given by
