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2 years ago

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A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later

. How
high is the cliff and how far from the base of the cliff did the diver hit the water?
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1 answer:

mart [117]2 years ago

8 0

Explanation:

It is given that,

The horizontal speed of a cliff diver, $v_x=3.6\ m/s$

It reaches the water below 2.00 s later, t = 2 s

Let $d_x$ is the distance where the diver hit the water. It can be calculated as follows :

$d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m$

Let $d_y$ is the height of the cliff. It can be calculated using second equation of motion as follows :

$d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m$

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

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For the car in this problem, the motion is:

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2 years ago

⦁ A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the c

Rudik [331]

Answer:

The value of temperature at compressor outlet = 543.43 K

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Pressure ratio ( $r_{p}$ )= 8 And specific heat ratio ( $\gamma$ ) = 1.4

Compressor inlet temperature ( $T_{1}$ ) = 300 K

Turbine inlet temperature ( $T_{3}$ ) = 1400 K

(a). Temperature ratio inside the compressor is given by

$\frac{T_{2} }{T_{1} }$ = $r_{p}\frac{\gamma - 1}{\gamma}$

Where $T_{1}$ & $T_{2}$ represents the compressor inlet and outlet temperature.

Put all the values in the given formula we get

⇒ $\frac{T_{2} }{300}$ = $8^{\frac{1.4 - 1}{1.4} }$

⇒ $\frac{T_{2} }{300}$ = 1.811

⇒ $T_{2}$ = 543.43 K

This is the value of temperature at compressor outlet.

The temperature ratio inside the turbine is given by

$\frac{T_{3} }{T_{4} }$ = $r_{p}\frac{\gamma - 1}{\gamma}$

Where $T_{3}$ & $T_{4}$ represents the turbine inlet & outlet temperatures.

Put all the values in the given formula we get

⇒ $\frac{1400}{T_{4} }$ = $8^{\frac{1.4 - 1}{1.4} }$

⇒ $T_{4}$ = 773.05 K

This is the temperature at turbine outlet.

(b). The work done by the turbine is given by the formula ( $W_{T}$ ) = $C_{p}$ ( $T_{3}$ - $T_{4}$ )

Put all the values in the above formula we get $W_{T}$ = 1.005 × (1400 - 773.05)

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And work done inside the pump is given by $W_{c}$ = $C_{p}$ ( $T_{2}$ - $T_{1}$ )

Put all the values in the above formula we get $W_{c}$ = 1.005 × (543.43 - 300)

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Put the values of $W_{c}$ & $W_{T}$ in above formula we get

⇒ $r_{b}$ = $\frac{244.65}{630.08}$

⇒ $r_{b}$ = 0.388

This is the value of back work ratio.

(C). Thermal Efficiency is given by E = 1 - $\frac{1}{r_{p} ^{\frac{\gamma - 1}{\gamma} } }$

⇒ E = 1 - $\frac{1}{8 ^{\frac{1.4 - 1}{1.4} } }$

⇒ E = 0.448

This is the value of efficiency of the Ideal Brayton cycle.

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