Answer:
The radius of the disc is 2.098 m.
(e) is correct option.
Explanation:
Given that,
Moment of inertia I = 12100 kg-m²
Mass of disc m = 5500 kg
Moment of inertia :
The moment of inertia is equal to the product of the mass and square of the radius.
The moment of inertia of the disc is given by
![I=\dfrac{mr^2}{2}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7Bmr%5E2%7D%7B2%7D)
Where, m = mass of disc
r = radius of the disc
Put the value into the formula
![12100=\dfrac{5500\times r^2}{2}](https://tex.z-dn.net/?f=12100%3D%5Cdfrac%7B5500%5Ctimes%20r%5E2%7D%7B2%7D)
![r=\sqrt{\dfrac{12100\times2}{5500}}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%5Cdfrac%7B12100%5Ctimes2%7D%7B5500%7D%7D)
![r= 2.098\ m](https://tex.z-dn.net/?f=r%3D%202.098%5C%20m)
Hence, The radius of the disc is 2.098 m.
Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Answer:
Energy lost due to friction is 22 J
Explanation:
Mass of the ball m = 4 kg
Initially velocity of ball v = 6 m/sec
So kinetic energy of the ball ![KE=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![KE=\frac{1}{2}\times 4\times 6^2=72J](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%204%5Ctimes%206%5E2%3D72J)
Now due to friction velocity decreases to 5 m/sec
Kinetic energy become
![KE=\frac{1}{2}\times 4\times 5^2=50J](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%204%5Ctimes%205%5E2%3D50J)
Therefore energy lost due to friction = 72 -50 = 22 J
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
I think you forgot to give the choices along with the question. I am answering the question based on my research and knowledge. <span>If a layer was deposited but does not appear in the rock record, the thing that happened is erosion. I hope that this is the ans wer that has actually come to your desired help.</span>