Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
Answer:

Explanation:
Given that
x= 150 ft

y= 14 ft
From the diagram

When ,x= 150 ft and y= 14 ft


z=150.74 ft

By differentiating with respect to time t


Here x is constant that is why


Now by putting the values in the above equation we get



Therefore the distance between balloon and observer increasing with 0.65 ft/s.
Answer:

Explanation:
For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

Where the r is the radius of merry-go-round and v is the tangential speed
but

So we have

Substitute the given values
So

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