Answer:
11.09 m/s
Explanation:
Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.
The parameters given are:
Initial velocity U = ?
Final velocity V = 9.6 m/s
Acceleration due to gravity g = 9.8m/s^2
Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0
Using third equation of motion
V^2 = U^2 - 2gH
0 = U^2 - 2 × 9.8H
U^2 = 19.6H ........ (1)
Using the formula again for one fourth of its maximum height
9.6^2 = U^2 - 2 × 9.8 × H/4
92.16 = U^2 - 19.6/4H
92.16 = U^2 - 4.9H
U^2 = 92.16 + 4.9H ...... (2)
Substitute U^2 in equation (1) into equation (2)
19.6H = 92.16 + 4.9H
Collect the like terms
19.6H - 4.9H = 92.16
14.7H = 92.16
H = 92.16/14.7
H = 6.269
Substitute H into equation 2
U^2 = 92.16 + 4.9( 6.269)
U^2 = 92.16 + 30.72
U^2 = 122.88
U = 11.09 m/s
Therefore, the initial velocity of the object is 11.09 m/s
Answer:
the answer is B. it's too easy
Answer:
m(P4) = 46.175 (grams)
m (KClO3) = 149 (grams)
Explanation:
1) n(P4) = n(P4O10);
m(P4)/M(P4) = m(P4O10)/M(P4O10);
m(P4) = M(P4)*m(P4O10)/M(P4O10)
= 123.90*105.8/283.89
= 46.175 (grams)
2) Analogously, 10n(P4O10) = 3n(KClO3)
m (KClO3) = 10M(KClO3)*m(P4O10)/3M(P4O10)
= 10*122.55*105.8/283.89/3
= 149 (grams).
Answer:
An aqueous stagnant layer that overlies the apical membrane and the subepithelial blood flow are potential barriers to the absorption of drugs that readily penetrate the absorbing cell of the epithelium. The apical, basal, and basement membranes are potential barriers to the absorption of less permeable drugs.
Glass and water are thicker and heavier than air. They are said to be 'denser' than air. What happens is that light slows down when it passes from the less dense air into the denser glass or water. This slowing down of the ray of light also causes the ray of light to change direction.
