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Answer:
The anchor points are 78.37 ft and 111.99 ft
Explanation:
If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then
a) Sine (α) = b/c it means opposite side/hypotenuse
b) Cosine (α)= a/c, it means adjacent side/hypotenuse
c) Tangent (α) = b/a opposite side/adjacent side.
Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:
a² + b² = c²
As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).
Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.
To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x
⇒ x = 78.37 ft.
And finally, to find y we can apply Pythagoras theorem
(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft
Summarizing, the anchor points are 78.37 ft and 111.99 ft
0.345 m.
<h3>Explanation</h3>The wavelength is the distance that the wave travels in each cycle. The wave travels 345 meters in each second. Let the wavelength of this wave be . That's the distance the wave travels in one cycle.
The frequency of the sound wave is 1 000 Hz, meaning that there are 1 000 cycles in each second. The wave travels a distance of 1 000 wavelengths in one second. That would be a distance of .
From the speed of the wave, the wave travels 345 meters in one second. In other words,
.
.
To generalize:
,
where
wavelength of the wave,
the speed of the wave, and
the frequency of the wave.
Answer:
0.66 m/s
Explanation:
This is an inelastic collision, for that reason the conservation of momentum law allows us to determine the final velocity of both, the bullet and block together after the collision:
(1) The potential energy at the top of the ball’s motion is 18 J.
(3) The kinetic energy increases as the potential energy decreases.
(4) The kinetic energy decreases as the potential energy increases.
(5) The total mechanical energy of the ball stays constant.
Explanation:
The total mechanical energy of the ball is equal to the sum of its kinetic energy (K, energy due to the motion) and its potential energy (U, energy due to the height of the ball). Mathematically:
In absence of friction, the mechanical energy of the ball is conserved, so in this case, it is always equal to 18 J. Let's now use this information to analyze each of the given statements:
(1) The potential energy at the top of the ball’s motion is 18 J. --> TRUE. In fact, at the top, the ball's speed becomes zero, so its kinetic energy is zero: K = 0. This means that all the mechanical energy of the ball is potential energy, therefore
E = U = 18 J
(2) The kinetic energy is less when the ball is thrown than when it is caught. --> FALSE. As we said, in absence of friction, the mechanical energy is conserved, therefore it always remains equal to 18 J.
(3) The kinetic energy increases as the potential energy decreases. --> TRUE. As we said, the sum of potential+kinetic energy remains constant:
E = K + U = 18 J
therefore, when the potential energy decreases, the kinetic energy increases.
(4)The kinetic energy decreases as the potential energy increases. --> TRUE. For the same reason described in (3).
(5)The total mechanical energy of the ball stays constant. --> TRUE. As we said at the beginning, the total mechanical energy is constant.
(6) The mechanical energy decreases as the ball moves up and increases as the ball comes down. --> FALSE. As we said, the mechanical energy remains constant, so it cannot change.
Learn more about kinetic and potential energy:
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