Answer:
Mas eu acho q é o empuxo será igual e a porção imersa do navio será maior.
Explanation: SE
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Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.
Answer:
x = 1,185 m
, t = 4/3 s
, F = - 4 N
Explanation:
For this exercise we use Newton's second law
F = m a = m dv /dt
β - α t = m dv / dt
dv = (β – α t) dt
We integrate
v = β t - ½ α t²
We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t
v-v₀ = β t - ½ α t²
the farthest point of the body is when v = v₀ = 0
0 = β t - ½ α t²
t = 2 β / α
t = 2 4/6
t = 4/3 s
Let's find the distance at this time
v = dx / dt
dx / dt = v₀ + β t - ½ α t2
dx = (v₀ + β t - ½ α t2) dt
We integrate
x = v₀ t + ½ β t - ½ 1/3 α t³
x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³
The body comes out of rest
x = 3.5556 - 2.37
x = 1,185 m
The value of force is
F = β - α t
F = 4 - 6 4/3
F = - 4 N
I found the answers here. Hope this helps you! https://1.cdn.edl.io/sJTle6yxt3qVq7jHfdHRZJ3Xogj7ps6swBO9umNcZ6PO3SMN.docx
