Answer:
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Answer:28.8 knots
Explanation:
The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.
First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.
The distance between them is described with Pythogorean theorem, which is
and when we replace the values A and D, we find Distance (D) to be 182 miles.
Now, let's make the notations clear. The velocity of A and B is notated as
and
. The rate of change of distance is also notated as
. Now, we have to find
from the Pythogorean theorem. If we derive the Pythogorean expression
, we would have:

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:
and the answer is 28.6 knots.
Answer:
0.304 m/s2
Explanation:
If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.
We can use Newton's 2nd law to calculate the net acceleration of the system

Answer:
96.7 s
Explanation:
Time of flight in projectile can be calculated thus:
T = 2 × u × sin ϴ/ g
Where;
T = time of flight (s)
u = initial velocity (m/s)
ϴ = Angle of projectile (°)
g = acceleration due to gravity (9.8m/s²)
Based on the provided information; u = 670m/s, ϴ = 45°
Hence, using T = 2.u.sin ϴ/ g
T = 2 × 670 × sin 45° ÷ 9.8
T = 1340 × 0.7071 ÷ 9.8
T = 947.52 ÷ 9.8
T = 96.68
T = 96.7s
O3 has molar mass of 48 g/mol
Therfore I mole weighs 48 grams
9.11 moles of ozone has a mass of 9.11 x 48grams = 437grams = 0.437kg