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Lelu [443]
2 years ago
10

A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that

is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
(a) 1.5 m/s
(b) 3.3 m/s
(c) 2.2 m/s
(d) 5.0 m/s
(e) 0.98 m/s
Physics
1 answer:
hram777 [196]2 years ago
8 0

Answer:

V_2=3.3m/s

Explanation:

From the question we are told that:

Distance d_1=1.4m

Tangential speed V=2.2m/s

Distance 2 d_2=2.1m

Generally the equation for Angular velocity is mathematically given by

 w=\frac{v}{r}

Therefore

 \frac{v_1}{r_1}=\frac{v_2}{r_2}

 V_2=\frac{2.2*2.1}{1.4}

 V_2=3.3m/s

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1.28 x 10^4 N

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4 years ago
Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be
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Answer:

The mass rate of the cooling water required is: 1'072988.5\frac{kg}{h}

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

{m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}

Where w refers to the cooling water and s to the steam flow. Reorganizing,

m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}

Write the difference of enthalpy for water as Cp (Tout-Tin):

m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, {h_{s}}^{out}=251.40\frac{kJ}{kg} and {h_{s}}^{in} can be calculated as:

{h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}.

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 \frac{kJ}{kgK}. If the average temperature is actually different, it won't mean a considerable mistake. Also we know that {T_{w}}^{out}-{T_{w}}^{in}\leq 10, so let's work with the limit case, which is {T_{w}}^{out}-{T_{w}}^{in}=10 to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}

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