Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.17 m on a
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Answer:
a) # buses = 7
Explanation:
For this exercise we use the kinematic equations, let's find the time it takes to reach the same height
y = t - ½ g t²
Let's decompose the speed, with trigonometry
v₀ₓ = v₀ cos θ
= v₀ sin θ
v₀ₓ = 40 cos 32
v₀ₓ = 33.9 m / s
= 40 sin32
= 21.2 m / s
When it arrives it is at the same initial height y = 0
0 = ( - ½ gt) t
That has two solutions
t = 0 when it comes out
t = 2 / g when it arrives
t = 2 21.2 /9.8
t = 4,326 s
We use the horizontal displacement equation
x = vox t
x = 33.9 4.326
x = 146.7 m
To find the number of buses we can use a direct proportions rule
# buses = 146.7 / 20
# buses = 7.3
# buses = 7
The distance of the seven buses is
L = 20 * 7 = 140 m
b) let's look for the scope for this jump
R = vo2 sin2T / g
R = 40 2 without 2 32 /9.8
R = 146.7 m
As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)
Well first we can tell that it will be very cold. For ex. Earth is fairly close to the sun. It gets pretty hot whenever the sun is out. Therefore we know being closer to the sun creates warmth on the planet. And as you can see on the picture of the solar system I attached, it seems as if the planet gets bigger the further away from the sun. So I would appose cold and big. <span />
