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15 days ago

7

A 1/10th scale model of an airplane is tested in a wind tunnel. The reynolds number of the model is the same as that of the full

scale airplane in air. The velocity in the wind tunnel is then

1 answer:

Yuliya22 [3.3K]15 days ago

7 0

Answer:

the airspeed of a full-sized aircraft

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Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.

inna [3103]

The heat required to raise the temperature of a substance by $\Delta T$ is represented by
$Q=m C_p \Delta T$
where m stands for the mass of the substance and $C_p$ indicates the specific heat of the substance. In this situation, we possess $m=100~g=0.1~Kg$ and $C_p=4.19~KJ/(Kg K)$ , the specific heat of water.
Consequently, we can ascertain the temperature rise $\Delta T$ :
$\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C$
Initially, the water's temperature was $25^{\circ}C$ , so the end temperature should be
$T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C$
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
$Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ$
where $C_L$ denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of $100^{\circ}C$ (the temperature at which the phase change begins)

4 0

1 month ago

A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab technician begins to supply heat to the container a

Maru [3345]

The response is 176 minutes. The translation of 456 g equals 0.456 kg. The specific heat of ice is 2093 J kg⁻¹, used to calculate heat required for a 25-degree rise, determined by mass multiplied by specific heat and temperature increase. The necessary calculations yield a total heat load of 176164 J. Finally, by dividing heat required by heat supply rate, we ascertain that it will take approximately 176.16 minutes.

4 0

21 day ago

A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What

kicyunya [3294]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = $60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s$

s = Distance traveled = $123\ ft=\dfrac{123}{3.281}=37.4885705578\ m$

$\theta$ = Incline = $26^{\circ}$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2$

Friction coefficient

$\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561$

$mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m$

The calculated stopping distance is 83.1946504051 m

6 0

2 months ago

A transverse wave is traveling from north to south. Which statement could be true for the motion of the wave particles in the me

Maru [3345]

Transverse waves propagate in a direction that is at right angles to the movement of the particles (or the medium involved). Hence, the particles would be shifting from east to west, which is perpendicular to the north-south direction of the wave.

3 0

29 days ago

Read 2 more answers

Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it

inna [3103]

Answer:

The cooker receives an average energy transfer rate of 1.80 kW.

Explanation:

Given that,

Pressure of the boiling water = 300 kPa

Mass of water = 3 kg

Duration = 30 min

Dryness fraction of the water = 0.5

What is the average energy transfer rate to the cooker?

We know that,

The specific enthalpy of vapor at 300 kPa is

$h_{f}=561.47\ kJ/kg$

$h_{fg}=2163.8\ kJ/kg$

We need to determine the initial state enthalpy of water

$h_{1}=h_{f}$

$h_{1}=561.47\ kJ/kg$

We must find the enthalpy of water in the final state

Using the enthalpy formula

$h_{2}=h_{f}+xh_{fg}$

Insert the value into the formula

$h_{2}=561.47+0.5\times2163.8$

$h_{2}=1643.37\ kJ/kg$

Next, we calculate the energy transfer rate to the cooker

Using the energy rate formula

$Q=\dfrac{m(h_{2}-h_{1})}{t}$

Substituting the value into the formula

$Q=\dfrac{3\times(1643.37-561.47)}{30\times60}$

$Q=1.80\ kW$

Therefore, the average energy transfer rate to the cooker stands at 1.80 kW.

3 0

1 month ago