Answer:
the terminal velocity of 14 nested coffee filters is 3.2 m/s
Explanation:
Given the data in the question;
we know that;
The terminal velocity is proportional to the square root of weight.
v ∝ √W
v = k√W
the proportionality constant depends upon the surface area and the density of the medium (like air). The coffee filters can be stacked such that the resulting area is roughly unchanged. So, the constant of proportionality k is also unchanged
v/√W = constant
v₂/√W₂ = v₁/√W₁
v₂ = v₁√(W₂ / W₁ )
given that;
v₁ = 0.856 m/s,
W₂ = 14W₁; meaning 14 coffee filters have 14 times the weight of a single coffee filter
so we substitute
v₂ = 0.856 √(14W₁ / W₁ )
v₂ = 0.856 √( 14( W₁/W₁)
v₂ = 0.856 √( 14(1)
v₂ = 0.856 √( 14 )
v₂ = 0.856 × 3.741657
v₂ = 3.2 m/s
Therefore, the terminal velocity of 14 nested coffee filters is 3.2 m/s
Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,

Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
Answer:
equilibrium position.
Explanation:
In simple harmonic motion , velocity v(t) is given by,
v(t) = -ω A sin(ωt + φ)
where
ω = angular velocity of the corresponding circular motion
A = amplitude
t = time
φ = the initial angle of the corresponding circular motion when the motion begin.
v (t) get maximized when sin value is maximized , i.e. sin
=1
The particle has maximum speed when it passes through the equilibrium position.
B) The amount of work done