Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
The correct answer is D. An average city
Explanation:
A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.
Answer: it can be considered a genetic mutation with a history of a Golden Retriever in their blood but it is very rare. and there our some black retrievers you can buy too. i hope i helped
Explanation:
Answer:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is
, and the steam properties as, A-4E:
![h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R](https://tex.z-dn.net/?f=h_%7Bfg%7D%3D1007.8Btu%2Flbm%2C%20s_%7Bfg%7D%3D1.6529Btu%2Flbm.R)
Using the energy balance for the system:
![\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s](https://tex.z-dn.net/?f=%5Cdot%20E_%7Bin%7D-%5Cdot%20E_%7Bout%7D%3D%5Cbigtriangleup%20%5Cdot%20E_%7Bsys%7D%3D0%5C%5C%5C%5C%5Cdot%20E_%7Bin%7D%3D%5Cdot%20E_%7Bout%7D%5C%5C%5C%5C%5Cdot%20Q_%7Bin%7D%2B%5Cdot%20m_%7Bcw%7Dh_1%3D%5Cdot%20m_%7Bcw%7Dh_2%5C%5C%5C%5C%5Cdot%20Q_%7Bin%7D%3D%5Cdot%20m_%7Bcw%7Dc_p%28T_%7Bout%7D-T_%7Bin%7D%29%5C%5C%5C%5C%5Cdot%20Q_%7Bin%7D%3D44%5Ctimes%201.0%5Ctimes%20%2873-60%29%3D572%5C%20Btu%2Fs)
Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:
![\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s](https://tex.z-dn.net/?f=%5Cdot%20m_%7Bsteam%7D%3D%5Cfrac%7B%5Cdot%20Q%7D%7Bh_%7Bfg%7D%7D%5C%5C%5C%5C%5Cdot%20m_%7Bsteam%7D%3D%5Cfrac%7B572%7D%7B1007.8%7D%3D0.5676lbm%2Fs)
Entropy generation in the heat exchanger could be defined using the entropy balance on the system:
![\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R](https://tex.z-dn.net/?f=%5Cdot%20S_%7Bin%7D-%5Cdot%20S_%7Bout%7D%2B%5Cdot%20S_%7Bgen%7D%3D%5Cbigtriangleup%20%5Cdot%20S_%7Bsys%7D%5C%5C%5C%5C%5Cdot%20m_1s_1%2B%5Cdot%20m_3s_3-%5Cdot%20m_2s_2-%5Cdot%20m_4s_4%2B%5Cdot%20S_%7Bgen%7D%3D0%5C%5C%5C%5C%5Cdot%20m_ws_1%2B%5Cdot%20m_ss_3-%5Cdot%20m_ws_2-%5Cdot%20m_ss_4%2B%5Cdot%20S_%7Bgen%7D%3D0%5C%5C%5C%5C%5Cdot%20S_%7Bgen%7D%3D%5Cdot%20m_w%28s_2-s_1%29%2B%5Cdot%20m_s%28s_4-s_3%29%5C%5C%5C%5C%5Cdot%20S_%7Bgen%7D%3D%5Cdot%20m%20c_p%20%5C%20In%28%5Cfrac%7BT_2%7D%7BT_1%7D%29-%5Cdot%20m_ss_%7Bfg%7D%5C%5C%5C%5C%5C%5C%5Cdot%20S_%7Bgen%7D%3D4.4%5Ctimes%201.0%5Ctimes%20%5C%20In%28%20%7B73%2B460%29%2F%2860%2B460%29%7D-0.5676%5Ctimes%201.6529%5C%5C%5C%5C%3D0.1483%5C%20Btu%2Fs.R)
Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R