Answer:
v = 26.7 mph
Explanation:
During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:
Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi
Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:
Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi
So, the total displacement during all the trip wil be:
Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi
So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:
v = Δx / Δt = 160 mi / 6 h = 26.7 mph
Your question seems to be incorrect. Please check below:
What force must the deltoid muscle provide to keep the arm in this position? By what factor does this force exceed the weight of the arm?<span>If you hold your arm outstretched with palm upward, as in (Figure 1) , the force to keep your arm from falling comes from your deltoid muscle. Assume that the arm has mass 4 kg and the distances and angles shown in (Figure 1) .
F=?
F/w= ?
The answer is </span><span>339 N</span><span>
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