Oxygen is obtained on a commercial scale through the liquefaction and distillation of ambient air at air separation plants. A secondpurification process may be necessary if ultra-high purity levels are required. High-purity oxygen can also be produced through the electrolysis of water.
Answer:
Image B
Explanation:
although I'm not exactly sure, i've recently gotten this question as well. but model B demonstrates the force- distance trade off because you can see how in that image them distance is increased in the force is decreased with the object being shorter. hopefully this helps in some way
Answer:
Explanation:
When the skier reaches the bottom of the slope , height lost by it
h = 50 sin32 m
= 26.5 m
potential energy lost
= mgh
Gain of kinetic energy
= 1/2 mv²
mgh = 1/2 mv²
v = √ 2gh
= √ (2x9.8 x 26.5)
= 22.8 m /s
b )
Let μ be the coefficient of kinetic friction required.
friction force acting
= μmg
work done by friction in displacement of d (40 m ) on horizontal surface
- μmg x d
This negative work will be equal to positive kinetic energy of the skier on horizontal surface .
= μmg x d = (1/2) m v²
μ = v² / (2 gd)
= 519.4 / (2 x 9.8 x 140 )
= .19
Answer:
2.85 m
Explanation:
From trigonometry,
Cosine = Adjacent/Hypotenuse
Assuming, The wall, the ladder and the ladder forms a right angle triangle as shown in fig 1, in the diagram attached below.
cos∅ = a/H....................... Equation 1
Where ∅ = Angle the ladder makes with the horizontal, a = The horizontal distance from the bottom of the ladder to the wall, H = The length of the ladder.
make a the subject of the equation
a = cos∅(H)..................... Equation 1
Given: ∅ = 68 °, H = 7.6 m.
Substitute into equation 2
a = cos(68)×7.6
a = 0.375×7.6
a = 2.85 m.
Hence the horizontal distance from the bottom of the ladder to the wall = 2.85 m
Answer:
8.3 x 10⁻⁷ C
Explanation:
Electric flux will enter the face at x=0 and exit at face x= 25 m
On the other faces , field lines are parallel so no flux will enter or exit .
Flux entering the face at x = 0
= electric field x face area
= 560 x 25 x 25 = 350000 weber
Flux exiting the face at x = 25
= 410 x 25 x25
= 256250 weber
Net flux exiting from cube ( closed face )
350000 - 256250 = 93750 web
Apply gauss'es theorem
Q / ε = Flux coming out
Q is charge inside the closed cube
Q / ε = 93750
Q = 8.85 x 10⁻¹² x 93750
= 8.3 x 10⁻⁷ C
