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Elan Coil
2 months ago
9

As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth

h1 = 42 mm and specific gravity S = 0.8. The reservoir is connected to a round tube of diameter D2 = 5 mm and oil rises in the tube to height h2. Find h2. Assume the oil in the tube is open to atmosphere and neglect the mass of the piston.
Engineering
1 answer:
mote1985 [299]2 months ago
7 0

Answer:

165 mm

Explanation:

The weight on the piston generates pressure on the oil using the equation:

p = f / A

Here, the force is represented by the mass's weight:

f = m * a

Where 'a' refers to the acceleration due to gravity.

A signifies the piston area, determined by:

A = π/4 * D1^2

Consequently, we have:

p = m * a / (π/4 * D1^2)

The height to which the oil will rise corresponds to that of a column that can exert equivalent pressure at its bottom:

p = f / A

The column's weight is given by:

f = m * a

The mass of this column is derived from its volume and specific gravity:

m = V * S

The volume can be calculated as the base area times the height:

V = A * h

Therefore:

p = A * h * S * a / A

We can simplify by eliminating the areas:

p = h * S * a

Next, we set pressures from both the piston and the column equal:

m * a / (π/4 * D1^2) = h * S * a

By simplifying further, we get:

m / (π/4 * D1^2) = h * S

Rearranging gives us:

h = m / (π/4 * D1^2 * S)

Here, 'h' denotes the height relative to the interface between the oil and the piston, with h1 being 42 mm. Therefore, the total height is expressed as:

h2 = h + h1

h2 = h1 + m / (π/4 * D1^2 * S)

h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm

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Answer:

t'_{1\2} = 6.6 sec

Explanation:

The half-life for the specified RC circuit can be expressed as

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.

Thus, the new half-life is

t'_{1\2} =R'Cln2

Now, divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

After substituting all values, we can calculate the revised half-life

t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
2 months ago
An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side
Mrrafil [318]

Answer:

Total bandwidth: 8 kHz

Explanation:

Data provided:

Transmitter frequency: 3.9 MHz

Modulation up to: 4 kHz

Solution:

For the upper side frequencies:

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Upper side frequencies = 3.904 MHz

For the lower side frequencies:

Lower side frequencies = 3.9 × 10^{6} - 4 × 10³

Lower side frequencies = 3.896 MHz

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6 0
2 months ago
Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe
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Flame-resistant clothing and aprons

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Workers involved with welding are generally mandated to wear flame-resistant clothing and aprons to shield them from various hazards, including heat, flames, burns, and potential radiation. In the context of welding, this gear protects individuals from flying sparks that can ignite and cause fires. Hence, such clothing helps to prevent accidents in these situations.

7 0
3 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>
6 0
2 months ago
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