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Lady_Fox [76]
3 years ago
6

A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom

thickness of glass). The glass has a density of 165 lb/ft3. The jar is placed in water with a density of 62.5 lb/ft3.
Assume the jar sits upright in the water without tipping over. How far will the empty jar sink into the water?

What is the volume of the glass shell of the jar? Precision 0.00

What is the weight of the jar? Precision 0.00


What is the weight of water the empty jar will displace? Precision 0.00

What is the volume of water the empty jar will displace? Precision 0.00

How far will the empty jar sink?
Engineering
1 answer:
swat323 years ago
6 0

Answer:

1. Volume of the glass shell (Vg) is simply volume of the empty part of the jar (Ve) subtracted from volume of the entire jar (Vj):

Vg = Vj - Ve

Volume is calculated as base (B) multiplied with height (h). Base of the jar is circle, so its surface is πr^2 (r being the radius).

However radius is different depending on the part of the jar; for empty part of the jar, inner radius is d = 3 in, for the whole jar it is inner radius plus thickness of the glass a = 3 + 3/16 = 3.1875 in.

We are also given height of the whole jar, h = 6 in, but height of the empty part is entire height minus thickness of the jar h' = 6 - 0.1875 = 5.8125 in.

Now, let's calculate:

Vj = πa^2 • h = 191.42 in^3

Ve = πd^2 • h' = 164.26 in^3

So, volume of the glass shell is Vj - Ve which is 27.16 in^3.

2. Mass of the glass jar is density of the glass multiplied with volume:

m = ρ • Vg

Density of the glass is given here in cubic feet so, first, we need to convert it to cubic inches, dividing it by 1728:

ρ = 165 lb/ft^3 / 1728 = 0.095 lb/in^3

So, mass of the jar is:

m = 0.095 lb/in^3 • 27.16 in^3 = 2.59 lb

5. To find weight and volume of the water displaced we first need to find how deep the jar sinks (H), because volume of the displaced water is equal to the volume of the jar submerged. Jar will sink until gravity force (pulling it down) and buoyancy force (pushing it up) become equal. Displaced water is πa^2 • H and the buoyancy is ρw • g • Vd (ρw is density of water which is 62.5 lb/ft^3 / 1728 = 0.036 lb/in^3, and Vd is displaced water).

So, buoyancy is:

B = ρw • g • πa^2 • H

We said that buoyancy must be equal to gravity:

B = m • g (m being mass of the jar). So:

ρw • g πa^2 • H = m • g

ρw • πa^2 • H = m

From this, we can find H:

H = m / ρw•πa^2

H = 2.25 inches

That means that the jar will sink 2.25 inches in the water.

3. Now, it's easy to find volume of displaced water. It's the same as the volume of the jar submerged:

Vd = πa^2 • H

Vd = 71.94 in^3

4. And finally, the weight of water is:

m = ρw • Vd

m = 0.036 lb/in^3 • 71.94 in^3

m = 2.59 lb

Of course, we see that the mass of the jar equals the mass of the displaced water. Taking this as a rule, this question could have been solved easier However I wanted to do it more detailed, to explain it more clearly

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Answer:

B.

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4 0
3 years ago
Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.
marin [14]

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30  + ( bicarbonate ion ) * 50/61

                                   = 15 * (50/30) + 120*( 50/61 )  = 123.361 mg/l  as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57  = 2.692*10^-5  moles/l

[ OH^- ]   = 2.692*10^-5  * 179/mole * 10^3 mg/g  = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3  = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

8 0
3 years ago
A solid shaft is subjected to an axial load P = 200 kN and a torque T = 1.5 kN.m. a) Determine the diameter of the shaft if the
Rom4ik [11]

Answer:

<em>a) 42 mm</em>

<em>b) 144.4 MPa</em>

<em></em>

Explanation:

Load P = 200 kN = 200 x 10^3 N

Torque T = 1.5 kN-m = 1.5 x 10^3 N-m

maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa

diameter of shaft d = ?

From T = τ * \frac{\pi }{16} * d^{3}

substituting values, we have

1.5 x 10^3 = 100 x 10^6 x \frac{3.142 }{16} x d^{3}

d^{3} = 7.638 x 10^-5

d = \sqrt[3]{7.638 * 10^-5} = 0.042 m = <em>42 mm</em>

b) Normal stress = P/A

where A is the area

A = \frac{\pi d^{2} }{4} = \frac{3.142*0.042^{2} }{4} = 1.385 x 10^-3

Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = <em>144.4 MPa</em>

7 0
4 years ago
Thank you,Thank you udahfsdf
Degger [83]

Answer: Um, a little feedback would help, cause I'm new but if you are thanking someone, let me know and i'll thank them for you, any other questions?

Explanation:

3 0
3 years ago
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical
Olin [163]

Answer:

The theoretical density for Niobium is 1.87 g/cm^3.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density  of the unit cell

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm

1 nm = 10^{-7} cm

On substituting all the given values , we will get the value of 'a'.

\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}

\rho =1.87 g/cm^3

The theoretical density for Niobium is 1.87 g/cm^3.

6 0
3 years ago
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