-- The object either left or crossed the starting line exactly at time=0 .
-- The object has been traveling at constant speed for all time that
we know about.
Long straight distance that a person can swim is 5.64 m.
<h3>What is the
Long straight distance?</h3>
The line that runs form one end of the circle to another is called the diameter of the circle. The pool is a circle according to the question and the long straight distance that a person can swim is the same of the diameter of the circular pool.
Now we have;
A = πr^2
A = area of pool
r = radius of pool
r = √A/ π
r = √25/3.142
r = 2.82m
Diameter of the circular pool = 2 r = 2 (2.82 cm) = 5.64 m
Learn more about circle: brainly.com/question/11833983
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Missing parts;
An ad for an above-ground pool states that it is 25 m2. From the ad, you can tell that the pool is a circle. If you swim from one point at the edge of the pool to another, along a straight line, what is the longest distance d you can swim? Express your answer in three significant figures.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Answer:
<h2>5.25 kg.m/s</h2>
Explanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 0.15 × 35
We have the final answer as
<h3>5.25 kg.m/s</h3>
Hope this helps you
Answer:
El peso del cartel es 397,97 N
Explanation:
La tensión dada en cada segmento del cable = 2000 N
El desplazamiento vertical del cable = 50 cm = 0,5 m
La distancia entre los polos = 10 m
La posición del letrero en el cable = En el medio = 5
El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °
El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero
El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N
El peso del signo = 397,97 N.
