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Juli2301
13 days ago
12

A tank contains 500 kg of a liquid whose specific gravity is 2. Determine the volume of the liquid in the tank.

Engineering
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In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe
Daniel [329]
The value obtained is 0.60. From the given conditions, we take k = 1.4 for air and r = 16. We know that for calculations involving diesel engine efficiency, we arrive at 0.60.
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1 month ago
The uniform dresser has a weight of 90 lb and rests on a tile floor for which the coefficient of static friction is 0.25. If the
Kisachek [356]

Answer:

a) F = 736.065\,lbf, b) \mu_{k} = 0.15

Explanation:

a) The uniform dresser can be modeled using specific equilibrium equations:

\Sigma F_{x} = F - \mu_{k}\cdot N = 0

\Sigma F_{y} = N-m\cdot g=0

Following some algebraic manipulations, the formulated equation is derived:

F = \mu_{k}\cdot m \cdot g

F = (0.25)\cdot (90\,lbm)\cdot (32.714\,\frac{ft}{s^{2}} )

F = 22.5\,lbf

b) Similarly, the man can be represented by a set of equilibrium equations:

\Sigma F_{x} = -F + \mu_{k}\cdot N = 0

\Sigma F_{y} = N-m\cdot g=0

After some algebraic changes, the expression for the coefficient of static friction comes out as:

\mu_{k} = \frac{F}{m\cdot g}

\mu_{k} = \frac{22.5\,lbf}{150\,lbf}

\mu_{k} = 0.15

3 0
2 months ago
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
mote1985 [299]

Response:

1. To find the volume of the glass shell (Vg), simply subtract the volume of the empty part of the jar (Ve) from the total volume of the jar (Vj):

Vg = Vj - Ve

Volume can be calculated by multiplying the base (B) with the height (h). The base of the jar is a circle, thus its area is πr^2 (where r indicates the radius).

The radius differs based on the jar's section: the inner radius for the empty part is d = 3 in, while for the total jar it includes the glass thickness a = 3 + 3/16 = 3.1875 in.

The height of the entire jar is given as h = 6 in, whereas for the empty portion, it's the total height minus the thickness of the glass h' = 6 - 0.1875 = 5.8125 in.

Now we can perform the calculations:

Vj = πa^2 • h = 191.42 in^3

Ve = πd^2 • h' = 164.26 in^3

Thus, the volume of the glass shell equals Vj - Ve, resulting in 27.16 in^3.

2. The mass of the glass jar can be determined by multiplying the density of the glass with the volume:

m = ρ • Vg

The glass density is provided in cubic feet, so we first convert it to cubic inches by dividing by 1728:

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

<ptherefore the="" mass="" of="" jar="" is:="">

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

5. To calculate the weight and volume of the displaced water, we first need to ascertain how deep the jar sinks (H), as the volume of displaced water equals the submerged volume of the jar. The jar will descend until the gravitational force downwards equals the buoyancy force upwards. The displaced water volume is πa^2 • H, and the buoyancy is calculated as ρw • g • Vd (where ρw is the density of water, defined as 62.5 lb/ft^3 / 1728 = 0.036 lb/in^3, and Vd is the displaced water volume).

Thus, the buoyancy can be represented as:

B = ρw • g • πa^2 • H

Setting buoyancy equal to gravity:

B = m • g (where m is the mass of the jar). Therefore, we have:

ρw • g • πa^2 • H = m • g

From this, simplifying gives:

ρw • πa^2 • H = m

We can derive H:

H = m / (ρw • πa^2)

H = 2.25 inches

This indicates the jar will sink 2.25 inches into the water.

3. Calculating the volume of displaced water is straightforward. It matches the volume of the submerged jar:

Vd = πa^2 • H

Vd = 71.94 in^3

4. Lastly, to determine the weight of the displaced water:

m = ρw • Vd

m = 0.036 lb/in^3 • 71.94 in^3

m = 2.59 lb

As evident, the mass of the jar aligns with the mass of the displaced water. Following this logic could have simplified our calculations, but I chose to elaborate for clarity.

</ptherefore>
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I2 + KOH = KIO3 + KI + H2O Marque la(s) respuesta(s) falsas: La suma de coeficientes mínimos del agua y el agente reductor es 6
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The incorrect statements include: - KOH is the reducing agent. - The total of transferred electrons alongside the minimum water coefficient rounds to 16. All other claims stand accurate. The false assertions point out that KOH does not function as the reducing agent, and the total of electrons and the water coefficient indeed equates to 13, rather than the stated 16.
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___________ is NOT a common injury that an automotive tech may experience at work.
Viktor [391]

Answer: The most frequently occurring injuries were sprains/strains, accounting for 39% of the total; lacerations comprised 22%, and contusions represented 15%. Almost half (49%) of the injuries led to one or more days of lost or restricted work; 25% resulted in 7 or more days lost or restricted.

Explanation:

Sprains/strains were the predominant injuries happening, making up 39% of all cases, while lacerations followed at 22% and contusions at 15%. Of these injuries, 49% caused employees to miss or face restricted workdays, with 25% leading to a minimum of 7 days lost or restricted.

7 0
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