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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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While at a concert you notice five people in the crowd headed in the same direction. Your tendency to group them is due to? *
Vlada [557]

Answer:

common fate

Explanation:

The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.

Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.

Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.

7 0
2 years ago
A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. If the
Bad White [126]

Answer:

=>> 167.3 kpa.

=>> 60° from horizontal face.

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;

=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "

=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."

The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).

magnitude of the stresses on the failure plane = 167.3 kpa.

The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.

y = 60 sin 60° = 30√3 = sheer stress.

the orientation of this plane with respect to the major principle stress plane.

Theta = 45 + 15 = 60°.

5 0
2 years ago
Q10. Select the correct option for the following questions – (10 points, 2 each) a. After an edge dislocation has passed through
Gnom [1K]

Answer:

a. True - Because the atomic arrangements of that region is disorderer because of the extra half plane atoms in between the line

b.Slip

C. Strength theoretical is greater than strength experimental

d. Shear stress

e. Highest linear density

4 0
2 years ago
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t
alukav5142 [94]

Answer:

  • hoop stress
  • longitudinal stress
  • material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = \frac{PD}{2T}     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = \frac{PD}{4T}       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

\frac{D}{T} ≥  20

insert this value into equation 1

αh = \frac{20 *30}{2}  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0
2 years ago
Problem 5) Water is pumped through a 60 m long, 0.3 m diameter pipe from a lower reservoir to a higher reservoir whose surface i
kap26 [50]

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2  =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV  =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f  = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

4 0
2 years ago
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