The difference between the measures of the intercepted arcs of the given circle is 45°.
A circle is defined as the set of points in a plane equidistant to each, such that it forms a closed two-dimensional figure, which is known as a circle.
- A line intersecting a circle at a minimum of two distinct points is known as a secant and the line touching the circle at only one point, is known as the tangent of a circle.
- The intercepted arc is the section of the circumference of a circle such that it is either encased by two chords or a line segment, meeting at a single point.
We are given that the angle subtended by secant and tangent, outside the circle is,
And angle measured inside the circle is,
So, the difference between the measures of the intercepted arcs is,
Thus, we can conclude that the difference between the measures of the intercepted arcs of the given circle is 45°.
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Explanation:
y = y₀ + v₀ t + ½ at²
For the first ball:
0 = h + v₀ t − 4.9t²
For the second ball:
0 = h − 4.9 (t−1)²
a) If h = 20.0, find v₀.
0 = 20 − 4.9 (t−1)²
t = 3.02 s
0 = 20 + v₀ (3.02) − 4.9 (3.02)²
v₀ = 8.18 m/s
Graph:
desmos.com/calculator/uk1wzkxybt
If v₀ is given, find h.
First, find t in terms of v₀:
h + v₀ t − 4.9t² = h − 4.9 (t−1)²
v₀ t − 4.9t² = -4.9 (t−1)²
v₀ t − 4.9t² = -4.9 (t² − 2t + 1)
v₀ t − 4.9t² = -4.9t² + 9.8t − 4.9
v₀ t = 9.8t − 4.9
(9.8 − v₀) t = 4.9
t = 4.9 / (9.8 − v₀)
Therefore:
h = 4.9 (4.9 / (9.8 − v₀) − 1)²
bi) If v₀ = 6.0 m/s:
h = 4.9 (1 / (9.8 − 6.0) − 1)²
h = 2.66 m
bii) If v₀ = 9.5 m/s:
h = 4.9 (1 / (9.8 − 9.5) − 1)²
h = 26.7 m
c) As found in part a, the time it takes for the first ball to land is:
t = 4.9 / (9.8 − v₀)
If v₀ is greater than 9.8 m/s, the time becomes negative, which isn't possible. Therefore, vmax = 9.8 m/s. At this speed, the ball would reach its highest point after 1 second, the same time that the second ball is dropped. Two balls dropped at the same time from different heights cannot land at the same time.
d) If v₀ is less than 4.9 m/s, the time for the first ball to land becomes less than 1 second. Which means it will have already landed before the second ball is dropped. Therefore, vmin = 4.9 m/s.
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by reducing the amount of energy required for the reaction to occur. other factors such as <span>catalysts' ability to speed a process are temperature, concentration </span>
