Answer:
(a) 693.12 torr
(b) 68.5 kilopascals
(c) 0.862 atmosphere
(d) 1.306 atmospheres
(e) 36.74 psi
Explanation:
(a) 0.912 atm = 0.912 atm × 760 torr/1 atm = 693.12 torr
(b) 0.685 bar = 0.685 bar × 100 kPa/1 bar = 68.5 kPa
(c) 655 mmHg = 655 mmHg × 1 atm/760 mmHg = 0.862 atm
(d) 1.323×10^5 Pa = 1.323×10^5 Pa × 1 atm/1.01325×10^5 Pa = 1.306 atm
(e) 2.50 atm = 2.50 atm × 14.696 psi/1 atm = 36.74 psi
Answer:
a = (v² – v₀²)/ 2(s – s₀)
Explanation:
v² = v₀² + 2a (s – s₀)
We can make 'a' the subject of the above expression as follow:
v² = v₀² + 2a (s – s₀)
Subtract v₀² from both side
v² – v₀² = v₀² + 2a (s – s₀) – v₀²
v² – v₀² = v₀² – v₀² + 2a (s – s₀)
v² – v₀² = 2a (s – s₀)
Divide both side by (s – s₀)
(v² – v₀²)/ (s – s₀) = 2a
Divide both side by 2
(v² – v₀²)/ (s – s₀) ÷ 2 = a
(v² – v₀²)/ (s – s₀) × 1/2 = a
(v² – v₀²)/ 2(s – s₀) = a
a = (v² – v₀²)/ 2(s – s₀)
Answer: The location where two air masses meet.
Explanation:
A front is a middle zone between two air masses. Each air mass exhibit different densities. The air masses also differ in humidity and temperature. The front is responsible for changing the climatic condition of a region.
Answer:
To have a positive job, the two vectors must have the same direction
Explanation:
Work is a scalar defined as the scalar product of two vectors, the froce and the displacement. To have a positive job, the two vectors must have
collinear and that their arrows point in the same direction.
You can also appreciate this from the work equation
W = F. r
bold indicate vector
W = F r cos θ
With θ the angle between force and displacement
The answer would be a nail rusting, but really it could be any of these.
