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3 years ago

What is the difference between density and specific gravity

1 answer:

4 0

Density is defined as mass per unit volume. It has the SI unit kg m-3 or Kg/m-3 and is an absolute quantity. Specific gravity is the ratio of a material's density with that of water at 4 °C (where it is most dense and is taken to have the value 999.974 kg·m-3). It is therefore a relative quantity with no units.

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A heavy dart and a light dart are launched vertically by identical ideal springs. Both springs were initially compressed by the

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Explanation:

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4 years ago

A 40 kg skier starts at the top of a 12 meter high slope. At the bottom, she is traveling 10 miles. How much energy does she los

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3 years ago

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Dwight pulls his sister in her wagon with a force of 65 N at an angle of 50.0º to the ground. What are the magnitudes of the hor

NeTakaya

Refer to the diagram shown.

F = 65 N, the force exerted by Dwight.

The horizontal component of the force is
65 cos(50°) = 41.8 N
The vertical component of the force is
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Answer:
Horizontal: 41.8 N
Vertical: 49.8 N

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4 years ago

Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

Rina8888 [55]

this may help you
As far as the field goes, the two charges opposite each other cancel!
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3 years ago

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Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

At State 1:

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

At State 1:

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

At State 1;

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$