I've found the complete form of this problem from another website which is shown in the attached picture. The equation is:
F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²
a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.
F = -25 + (1.23×10⁵)(
22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)²
F = -69.62 N
From Newton's second law, F = ma.
-69.62 = (165 g)(1 kg/1000 g)(a)
Solving for a,
a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0)
Solving for v,
<em>v = 9.45 m/s</em>b.) To solve for the distance, the formula is:
d = v₀t + 1/2(a)(t²)
Let's use the absolute value of a because distance is always positive.
d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)²
<em>d = 0.106 m</em>
Blood is a buffer solution of bicarbonate (HCO₃⁻) and carbon dioxide (CO₂). The Henderson equation which relates the concentration of HCO₃⁻ and CO₂ is given below:
=
.
Respiring cell releases CO₂ in blood stream and that CO₂ on reaction with water molecule produces H₂CO₃ which is a weak base and its conjugate base is HCO₃⁻.
CO₂ + 2H₂O⇄ HCO₃⁻ + H₃O⁺
pH of the buffer solution (the blood) depends only on the ratio of the amount of CO₂ to the amount of HCO³⁻. [So, due to respiration produced CO₂ will get dissolved in water and favours the equilibrium towards forward direction. Then immediately HCO₃⁻ reacts with HCO₃⁻ and starts producing CO₂.]This ratio remains relatively constant because the concentrations HCO3- and CO2 are very large compared to the amount of CO₂ produced to the blood from respiring cells. So,
of blood does not change.
Answer:
A. It has a partial negative charge on oxygen and a partial positive charge on hydrogen.
Explanation:
There is an electronegativity difference between O and H atoms. The electronegativity value of O and H are 3.5 and 2.1 respectively. So the electronegativity different of the O-H bond is 3.5- 2.1 = 1.4 thus making the O-H bond polar. O being the more electronegative bond acquires a partial negative charge and H acquires a partial positive charge.
The wall would absorb extra heat during the day when the sun is out, then release the heat back into the room when the sun goes down.