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3 years ago

A cylindrical metal can is to have no lid. It is to have a volume of 125π in3. What height minimizes the amount of metal used?

Physics

1 answer:

Alex Ar [27]3 years ago

6 0

Answer:

Minimum height of metal = 5 inches

Explanation:

Volume of the cylindrical metal = πR²H = 125π

cancelling out π on both sides

R²H = 125

Hence it can be deduced that R² = 25 and H = 5

Hence minimum height of metal = 5 inches

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In order to protect underground pipelines and storage tanks made of iron, the iron is made the cathode of a voltaic cell through

emmasim [6.3K]

The answer would be that the process to protect the underground pipelines and storage tanks is cathodic protection.

3 0

3 years ago

A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H

Reika [66]

Answer:

The answer to your question is:

a) t1 = 2.99 s ≈ 3 s

b) vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ? time to reach the ground

vf = ? final speed

a) h = vot + (1/2)gt²

74 = 10t + (1/2)9.81t²

4.9t² + 10t -74 = 0 solve by using quadratic formula

t = (-b ± √ (b² -4ac) / 2a

t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

t = (-10 ± √ 1550.4 ) / 9.81

t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81

t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81

t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are

no negative times.

b) Formula vf = vo + gt

vf = 10 + (9.81)(3)

vf = 10 + 29.43

vf = 39.43 m/s

4 0

2 years ago

A boat takes 3.0 hours to travel 50 km down a river, then 5.4 hours to return. Determine the speed of the water in the river.

Nutka1998 [239]

Answer:

3.7 km/h

Explanation:

Let's call v the proper speed of the boat and v' the speed of the water in the river.

When the boat travels in the direction of the current, the speed of the boat is:

v + v'

And it covers 50 km in 3 h, so we can write

$v+v' = \frac{50 km}{3 h}=16.7 km/h$ (1)

When the boat travels in the opposite direction, the speed of the boat is

v - v'

And it covers 50 km in 5.4 h, so

$v-v'=\frac{50 km}{5.4 h}=9.3 km/h$ (2)

So we have a system of two equations: by solving them simultaneously, we find the value of v and v':

$v+v'=16.7 \\v-v'=9.3$

Subtracting the second equation from the first one we get:

$(v+v')-(v-v')=16.7-9.3\\2v'=7.4\\v'=3.7$

So, the speed of the water is 3.7 km/h.

5 0

3 years ago

A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the

fredd [130]

Answer:

t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material = 1.375

we know the equation

c = ν λ

where ν is the frequency of the wave

c is the speed of light

$\nu= \dfrac{c}{\nu\lambda}$

$\nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}$

ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

$t = \dfrac{\lambda}{4\mu_{material}}$

$t = \dfrac{522 \times 10^{-9}}{4\times 1.375}$

t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm

7 0

3 years ago

A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100

Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

$\rho$ = Density of fluid

v = Fluid velocity

m = Mass = $\rho Al$

Centripetal force is given by

$F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L$

Pressure is given by

$P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s$

The angular speed of the tube is 31.321 rad/s

5 0

3 years ago