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igor_vitrenko [27]
3 years ago
15

An athlete runs 200 m by running 100 m in a straight line and then turning around and running 100m back to the start point.

Physics
1 answer:
siniylev [52]3 years ago
6 0

Answer:

o

Explanation:

The athlete ran a total distance of zero because they ran 100m forward then turned around so they went back to their starting position

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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
4 years ago
The angle of incidence always equals what?
Ratling [72]

Answer:

Angle Of Incidence Formula The angle of incidence is equal to the reflected angle through the law of reflection. The angle of incidence and the angle of reflection is always equal, and they are both on the same plane along with the normal. 2,44,451

Explanation:

that could help you with work.

4 0
3 years ago
Read 2 more answers
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on
disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

V = V_1 + V_2 + V_3

here we have potential due to 50 nC placed at y = 6 m

V_1 = \frac{kQ}{r}

V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_1 = 45 Volts

Now potential due to -80 nC charge placed at x = -4

V_2 = \frac{kQ}{r}

V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}

V_2 = -60 Volts

Now potential due to 70 nC placed at y = -6 m

V_3 = \frac{kQ}{r}

V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_3 = 63 Volts

Now total potential at this point is given as

V = 45 - 60 + 63 = 48 Volts

3 0
3 years ago
Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation, in kW, from 1
Leya [2.2K]

Answer:

1.5 x 10⁵ W

Explanation:

A = Area of the fresh lava = 1.02 m²

T = Temperature of fresh lava = 1000 °C = 1000 + 273 = 1273 K

T₀ = Temperature of surrounding = 25.3 °C = 25.3 + 273 = 298.3 K

ε = emissivity of the lava = 0.97

σ = stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Rate of  transfer is given as

E = σ ε A (T⁴ - T₀⁴)

E = (5.67 x 10⁻⁸) (0.97) (1.02) ((1273)⁴ - (298.3)⁴)

E = 1.5 x 10⁵ W

3 0
3 years ago
A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it
mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm

r = 0.145\ cm = 0.145\times 10^{-2}\ m

Charge ,

Q = 30.0\ pC=30\times 10^{-12}

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}

Where,  

V= Electric potential,  

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge  

r = Radius  

Substituting the values we get

V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}

V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0
3 years ago
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