Answer:
s_max = 0.8394m
Explanation:
From equilibrium of block, N = W = mg
Frictional force = μ_k•N = μ_k•mg
Since μ_k = 0.3,then F = 0.3mg
To determine the velocity of Block A just before collision, let's apply the principle of work and energy;
T1 + ΣU_1-2 = T2
So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²
Plugging in the relevant values to get ;
(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²
750 - 176.58 = 7.5(v_a1)²
v_a1 = 8.744 m/s
Using law of conservation of momentum;
Σ(m1v1) = Σ(m2v2)
Thus,
m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2
Thus;
15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)
Divide through by 5;
3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)
Thus,
3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)
Coefficient of restitution has a formula;
e = (v_b2 - v_a2)/(v_a1 - v_b1)
From the question, e = 0.6.
Thus;
0.6 = (v_b2 - v_a2)/(8.744 - 0)
0.6 x 8.744 = (v_b2 - v_a2)
(v_b2 - v_a2) = 5.246 - - - (eq2)
Solving eq(1) and 2 simultaneously, we have;
v_b2 = 8.394 m/s
v_a2 = 3.148 m/s
Now, to find maximum compression, let's apply conservation of energy on block B;
T1 + V1 = T2 + V2
Thus,
(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²
(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²
500(s_max)² = 352.29618
(s_max)² = 352.29618/500
(s_max)² = 0.7046
s_max = 0.8394m
Answer: hello your question is incomplete below is the missing part
question :Determine the temperature of the cooled water exiting the cooling tower
answer : T = 43.477° C
Explanation:
Temp of water at exit = 45°C
mass flow rate of cooling tower = 15,000 kg/s
Temp of makeup water = 20°C
Assuming an atmospheric pressure of = 101.3 kPa
Determine temperature of the cooled water exiting the cooling tower
Water entering cooling tower at 45°C
Given that Latent heat of water at 45°C = 43.13 KJ/mol
Cp(wet air) = 1.005+ 1.884(y1)
where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273
Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C
First step : calculate the value of Q
Q = m*Cp*(ΔT) + W(latent heat)
Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)
Q = 95935.8547 KJ/s
Given that mass rate of water = 15000 kg/s
Hence the temperature of the cooled water can be calculated using the equation below
Q = m*Cp*∆T
Cp(water) = 4.2 KJ/Kg°C
95935.8547 = (15000)*(4.2)*(45 - T)
( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C
Answer:
See explaination
Explanation:
See attachment for the detailed step by step solution of the given problem.
The way a programmer describe a pre-emptive dialog by purely graphical means is; by producing a window that covers the entire screen to make it the currently selected window.
In a graphics - based interaction, it is supposed that the user can only interact with parts of the system that are visible. However, In a windowing system, the user can only direct input to a single window that was currently selected and the way to change that selected window is to indicate with some gesture within that window.
Finally, to create a pre-emptive dialog, the system would do so through the production of a window that covers the entire screen to make it the currently selected window. Thereafter, all user input would be directed to that window and the user would have no means of selecting any other window. Then the covering window will now pre-empt any other user action with the exception of that which it is defined to support.
Read more about dialogue at; brainly.com/question/5962406
Answer:
In an illustrations based communication, it is expected that the client can just associate with parts of the framework that are obvious. In a windowing framework, for instance, the client can just direct contribution to a solitary, at present chosen window, and the main methods for changing the chose window would be by demonstrating with some signal inside that window. To make a preemptive exchange, the framework can create a window that covers the whole screen and make it the right now chosen window. All client information would then be coordinated to that window and the client would have no methods for choosing another window. The 'covering' window in this way preempts some other client activity with the exception of that which it is characterized to help
Answer:
The solution is given in the attachments.
The four relevant pressures in a Rankine cycle with one stage of reheat are P1, P2, P3, and P4.
For a Rankine cycle with one stage of reheat between turbines, there are typically four relevant pressures:
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