A toy helicopter takes off and moves 6 m up and then 1 m back down. What

What terms are needed to completely describe velocity?

kotegsom [21]

In order to completely describe a velocity,
you need a speed and a direction.

4 0

3 years ago

To prevent accidental electric shock, most electrical equipment is A. painted. B. isolated. C. coated. D. grounded.

Nesterboy [21]

D. It is Grounded below ground

4 0

3 years ago

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Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

5 0

3 years ago

What layer of the Earth is the second layer from the surface?

Fantom [35]

Answer: The Mantle

Explanation: The second layer on earth from the surface should be the mantle

3 0

3 years ago

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Calculate the power transfer when a pump moves 50kg of water through a vertical height of 8meters in 5seconds (take g =10meters

Vika [28.1K]

Hi there!

Recall the following relationships:

$W = \Delta U = mg\Delta h\\\\P = \frac{W}{t}$

W = Work (J)

U = Potential Energy (J)

m = mass (kg)

g = acceleration due to gravity (9.8 m/s²)

h = height (m)

t = time (s)

Begin by calculating the work:

$W = 50(10)(8) = 4000 J$

Now, divide by the time to solve for power:

$P = \frac{4000}{5} = \boxed{800W}$

**W is the unit for power (Watts). Be careful not to get the two confused.

4 0

2 years ago