Answer:
C. Oxygen combines with carbon dioxide
Explanation:
B i o l o g y
Also, oxygen is a reactant and carbon dioxide is a product of cellular respiration that does not combine during this process
Hope it helps
Answer:
Correct answer: 11. Total distance d = 200m ; 12. Vav = 3.63m/s ;
13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;
15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s
Explanation:
The whole movement can be divided into four stages.
In the first stage the subject moves 60m in a positive direction for 10s,
in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.
11. Total distance = 60 + 0 + 100 + 40 = 200m
12. The formula for calculating the average speed (velocity) is
Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)
Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s
13. The movement started from the origin and ended at the origin
Total displacement is zero meters.
14. The speed between 10s and 15s is zero, because he did not move.
15. V₃ = S₃/t₃ = 100/25 = 4 m/s
16. V₁ = S₁/t₁ = 60/10 = 6 m/s and V₄ = S₄/t₄ = 40/15 = 2.67 m/s
V₁ > V₄
God is with you!!!
Answer:
Explanation:
When the spring is compressed by .80 m , restoring force by spring on block
= 130 x .80
= 104 N , acting away from wall
External force = 82 N , acting towards wall
Force of friction acting towards wall = μmg
= .4 x 4 x 9.8
= 15.68 N
Net force away from wall
= 104 -15.68 - 82
= 6.32 N
Acceleration
= 6.32 / 4
= 1.58 m / s²
It will be away from wall
Energy released by compressed spring = 1/2 k x²
= .5 x 130 x .8²
= 41.6 J
Energy lost in friction
= μmg x .8
= .4 x 4 x 9.8 x .8
= 12.544 J
Energy available to block
= 41.6 - 12.544 J
= 29 J
Kinetic energy of block = 29
1/2 x 4 x v² = 29
v = 3.8 m / s
This will b speed of block as soon as spring relaxes. (x = 0 )
Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.
