Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
Given parameters:
Mass on earth = 50kg
Unknown:
Mass on planet Xenon = ?
Weight on planet Xenon = ?
Mass is the amount of matter contained in a particular substance.
Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.
Weight = mass x acceleration due to gravity
Planet Xenon has half the gravitational force of Earth.
This translated gives 
= 4.9m/s²
Now, mass is always the same every where if the amount of matter in a substance does not change.
In this problem, mass = 50kg on planet xenon.
Weight = mass x acceleration due to gravity = 50 x 4.9 = 245N
The weight on Xenon is 245N and the mass is 50kg
Gravity is the attraction of every body to every other body due to the masses of each body. The larger the mass, the greater the force. It also depends on the distances: the closer the bodies, the greater the force. Gravity is directed toward the center of a body, and the distance is measured from the center.
When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.
It uses electromagnetic radiation waves to enable military communications, navigation, radar, nonintrusive inspection of aircraft, and other equipment. Hope this helps.
Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
