Fundamentals Of Thermodynamics
10th Edition
ISBN: 9781119494966
Author: Borgnakke, C. (claus), Sonntag, Richard Edwin, Author.
Publisher: Wiley,
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Chapter 6, Problem 6.10P
Why do we write
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In the Figure 1, if Q= 40 N, P = 50 N, and B =
145°, then R is equal to:
Figure 1:
R
B.
O 80 N
85 N
О 90 N
95 N
1
1
ӘР
2. Given: P = 2(R2) (S3) (T3) + (R3) (S)sin2T, what is әт
y
O, = 0, cos²0 + o, sin²0 + 2 Txy sin 0 cos 0
15.00
Tnt = -(0x - Oy) sin 0 cos 0 + txy( cos²0- sin?0)
These equations may also be written as:
Ox +Oy
Ox- Oy
On =
cos 20 + Txy sin 20
2
+
12.00
2
130°
Ox- Ov
10.00
Tnt =
sin 20 + Txy cos 20
10.00
12.00
On
Both the xyz and ntz
coordinate systems are
right-handed systems.
15.00
Tnt
The z axis is directed
toward the viewer.
1st 2nd
3rd
Enter
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1
Chapter 6 Solutions
Fundamentals Of Thermodynamics
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- 11. Athermally insulated battery is being discharged at atmospheric pressure and constant volume. During a 1 hour test it is found that a current of 50 A and 2 V flows while the temperature increases from 20°C to 32.5°C. Find the change in internal energy of the cell during the period of operation. [Ans.-36 x 104 J]arrow_forwardIn the Figure 1, if Q= 40 N, P = 50 N, and B = 145°, then R is equal to: * Figure 1: R B. O 80 N 85 N O 90 N 95 Narrow_forwardIn the Figure 1, if Q= 40 N, P = 50 N, and B = 145°, then R is equal to (in N): Figure 1: с R A Q B Parrow_forward
- Figure P D A B C 0arrow_forwardat atmospheric pressure, water melts as the temperature increases from 270K to 280K because a. the chemical potential of the liquid is lower than that of the gas b. the liquid and solid form a eutectic mixture c. the chemical potential of the liquid and solid become equal d. the chemical potential of the liquid is higher than that of the gasarrow_forwardIn the Figure 1, if Q= 40 N, R = 100 N, and B = 145°, then C is equal to: %3D %3D %3D Figure 1: R B. 10° 11° O 12° О 13°arrow_forward
- y On = 0, cos²0 + o, sin²0 + 2 Txy sin 0 cos 0 15.00 Tnt = -(0x - Oy) sin 0 cos 0 + txy( cos²0- sin?0) These equations may also be written as: Ox+Oy On Ox- Oy cos 20 + Txy sin 20 2 8.00 2 Ox- Oy Tnt sin 20 + Txy Cos 20 2 = - 10.00 10.00 55° 8.00 On Both the xyz and ntz coordinate systems are right-handed systems. The z axis is directed 15.00 Tnt toward the viewer. 1st 2nd 3rd Enter solved problems counter attemptarrow_forwardQuestion 1. Identify the correct description for the formula f'(t)h-¹(f(t+h) − f(t − h)) from the following options: FFD1: forward finite difference with stepsize h for the first derivative of f at t BFD1: backward finite difference with stepsize h for the first derivative of f at t CFD1: central finite difference with stepsize h for the first derivative of f at t CFD2: central finite difference with stepsize h for the second derivative of f at t None of the Above Make your selection here: Selectarrow_forwardFind the general solution toy′′−y′+ 9y= 3 sin(3t)using the method of undetermined coefficients.arrow_forward
- The Cyclic integral of a property is always zero. Select one: O True O Falsearrow_forwardA higher-order differential equation given as y" + p(x)y' g (x)y = O is a homogenous equation, even if p(x) = 0. A higher-order differential equation given as y" + p(x)y' + q(x)y = 0 is also a homogenous equation, even if g(x) = 0. O Only first statement is true. O Only second statement is true. O Both first and second statements are true. O Both first and second statements are false.arrow_forwardX =300 b/5t A 60° 45°arrow_forward
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