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Lady_Fox [76]
2 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:
swat322 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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An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectivel
Tju [1.3M]

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

3 0
2 years ago
The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc
Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

\theta = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑f_{y} = 0, ∑M = 0

If ∑M_{A} = 0

⇒F_{BC}×0.9 - P × 0.6 = 0

⇒F_{BC}×3 - P × 2 = 0

⇒F_{BC} = \frac{2P}{3}

If ∑M_{B} = 0

⇒F_{AD}×0.9 = P × 0.3

⇒F_{AD} ×3 = P

⇒F_{AD} = \frac{P}{3}

Now,

Area, A = \frac{\pi }{4} X (0.012)^{2} = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , \delta = \frac{P l}{A E}

Now,

\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 9.1626 × 10⁻⁹ P

\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9}  } = 1.83253 × 10⁻⁸ P

Given that,

\theta = 0.015°

⇒\theta = 2.618 × 10⁻⁴ rad

So,

\theta =  \frac{\delta_{BC} - \delta_{AD}}{0.9}

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0
2 years ago
The legend that Benjamin Franklin flew a kite as a storm approached is only a legend—he was neither stupid nor suicidal. Suppose
Delicious77 [7]

Answer: 0.93 mA

Explanation:

In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.

We can express the resistance as follows:

R = ρ.L/A

In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:

A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2

Replacing by the values, we get R as follows:

R = 1.4 1010 Ω

Applying Ohm’s Law, and solving for the current I:

I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA

7 0
2 years ago
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

Otto cycle T-S diagram

T₂ = 288.706*6.25^{0.393} = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}}  \right )^{K-1}

T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0
2 years ago
A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. a)
maks197457 [2]

Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars

Explanation :

A)

Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}

= 3780kJ

And 1 hour = 3600s

Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W

B)

At 15km/hour a 15km run takes 1 hour.

1 hour is 3600s and the runner burns 1050 joule per second.

Energy used in 1 hour = 3600 x 1050 J/s

= 3780000 J or 3.78MJ

C)

1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km

15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ

Finally,

1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ

This means that the runner needs 5320/1008 = 5.3 bars

7 0
2 years ago
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