What information you would gain by watching the video. Plz Help due today.

1. Two technicians are discussing tire rotation. Technician A says that you always follow the tire-rotation procedure outlined i

siniylev [52]

Answer:

don't know

Explanation:

huhuh

8 0

3 years ago

A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug

Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

$\rho = \frac{m}{V}$

Where

m = Mass

V = Volume

For state one we know that

$\rho_1 = \frac{m_1}{V}$

$m_1 = \rho_1 V$

$m_1 = 1.18*1$

$m_1 = 1.18Kg$

For state two we have to

$\rho_2 = \frac{m_2}{V}$

$m_2 = \rho_2 V$

$m_1 = 7.2*1$

$m_1 = 7.2Kg$

Therefore the total change of mass would be

$\Delta m = m_2-m_1$

$\Delta m = 7.2-1.18$

$\Delta m = 6.02Kg$

Therefore the mass of air that has entered to the tank is 6.02Kg

5 0

3 years ago

In a website browser address bar, what does “www” stand for?

Ludmilka [50]

Answer:

www stands for world wide web

Explanation:

It will really help you thank you.

3 0

3 years ago

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number ( $Re$ ) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

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

where $\rho$ is the density of the fluid, $\mu$ is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

$v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s$

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

$D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm$

8 0

3 years ago

A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is th

tigry1 [53]

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)

u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

0.014m³ / 0.2066 m³/kg

m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

u₁ = 419.06 kj / kg + 0.123 . 2087.0 kj/kg

u₁ = 675.76 kj/kg

For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ = 0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524 . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

7 0

3 years ago

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