Answer:
Gravitational; strongest; facing; closer; near side; toward.
Explanation:
The gravitational attraction between the Earth and the moon is strongest on the side of the Earth that happens to be facing the moon, simply because it is closer. This attraction causes the water on this “near side” of Earth to be pulled toward the moon. These forces of attraction and inertia tends to keep the water in place and consequently, leads to a bulge of water on the near side with respect to the moon.
Also, you should note that what is responsible for the moon being in orbit around the Earth is the gravitational force of attraction between the two planetary bodies (Earth and Moon).
Answer:
Modulus of resilience will be ![3216942.308j/m^3](https://tex.z-dn.net/?f=3216942.308j%2Fm%5E3)
Explanation:
We have given yield strength ![\sigma _y=818MPa](https://tex.z-dn.net/?f=%5Csigma%20_y%3D818MPa)
Elastic modulus E = 104 GPa
We have to find the modulus
Modulus of resilience is given by
Modulus of resilience
, here
is yield strength and E is elastic modulus
Modulus of resilience
Answer:
The original length of the specimen ![l_{o} = 104.7 mm](https://tex.z-dn.net/?f=l_%7Bo%7D%20%3D%20104.7%20mm)
Explanation:
Original diameter
= 30 mm
Final diameter
= 30.04 mm
Change in diameter Δd = 0.04 mm
Final length
= 105.20 mm
Elastic modulus E = 65.5 G pa = 65.5 ×
M pa
Shear modulus G = 25.4 G pa = 25.4 ×
M pa
We know that the relation between the shear modulus & elastic modulus is given by
![G = \frac{E}{2(1 + \mu)}](https://tex.z-dn.net/?f=G%20%3D%20%5Cfrac%7BE%7D%7B2%281%20%2B%20%5Cmu%29%7D)
![25.5 = \frac{65.5}{2 (1 + \mu)}](https://tex.z-dn.net/?f=25.5%20%3D%20%5Cfrac%7B65.5%7D%7B2%20%281%20%2B%20%5Cmu%29%7D)
![\mu = 0.28](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.28)
This is the value of possion's ratio.
We know that the possion's ratio is given by
![\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B%5Cfrac%7B0.04%7D%7B30%7D%20%7D%7B%5Cfrac%7Bchange%20%5C%20in%20%5C%20length%7D%7Bl_%7Bo%7D%20%7D%20%7D)
![{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}](https://tex.z-dn.net/?f=%7B%5Cfrac%7Bchange%20%5C%20in%20%5C%20length%7D%7Bl_%7Bo%7D%20%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B0.04%7D%7B30%7D%20%7D%7B0.28%7D)
0.00476
![\frac{l_{1} - l_{o} }{l_{o} } = 0.00476](https://tex.z-dn.net/?f=%5Cfrac%7Bl_%7B1%7D%20-%20l_%7Bo%7D%20%20%7D%7Bl_%7Bo%7D%20%20%7D%20%3D%200.00476)
Final length
= 105.2 m
Original length
![l_{o} = \frac{105.2}{1.00476}](https://tex.z-dn.net/?f=l_%7Bo%7D%20%3D%20%5Cfrac%7B105.2%7D%7B1.00476%7D)
![l_{o} = 104.7 mm](https://tex.z-dn.net/?f=l_%7Bo%7D%20%3D%20104.7%20mm)
This is the original length of the specimen.
Answer:
1.96 kg/s.
Explanation:
So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;
=> Superheated water vapor at a pressure = 20 MPa,
=> temperature = 500°C,
=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."
=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."
K1 = 3241.18, k2 = 93.28 and 2725.47.
Therefore, m1 + m2= m3.
10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.
=> 1.96 kg/s.