Answer:
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Let us say that x is the cut that we will make on the
sides to make a box, therefore the new dimensions are:
l = 15 – 2x
w = 8 – 2x
It is 2x since we cut on two sides.
We know that volume is:
V = l w x
V = (15 – 2x) (8 – 2x) x
V = 120x – 30x^2 – 16x^2 + 4x^3
V = 120x – 46x^2 + 4x^3
Taking the 1st derivative:
dV/dx = 120 – 92x + 12x^2
Set dV/dx = 0 to get maxima:
120 – 92x + 12x^2 = 0
Divide by 12:
x^2 – (92/12)x + 10 = 0
(x – (92/24))^2 = -10 + (92/24)^2
x - 92/24 = ±2.17
x = 1.66, 6
We cannot have x = 6 because that will make our w
negative, so:
x = 1.66 inches
So the largest volume is:
V = 120x – 46x^2 + 4x^3
V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3
V = 90.74 cubic inches
Answer:
105.8 m
46 m/s
Explanation:
From the time the rocket is launched to the time it reaches its maximum height:
v = 0 m/s
a = -10 m/s²
t = 9.2 s / 2 = 4.6 s
Find: Δy and v₀
Δy = vt − ½ at²
Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²
Δy = 105.8 m
v = at + v₀
0 m/s = (-10 m/s²) (4.6 s) + v₀
v₀ = 46 m/s
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Answer:
the volume is 0.253 cm³
Explanation:
The pressure underwater is related with the pressure in the surface through Pascal's law:
P(h)= Po + ρgh
where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)
replacing values
P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa
Also assuming that the bubble behaves as an ideal gas
PV=nRT
where
P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature
therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have
at the surface) PoVo=nRTo
at the depth h) PV=nRT
dividing both equations
(P/Po)(V/Vo)=(T/To)
or
V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³
V = 0.253 cm³