1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anna007 [38]
3 years ago
5

"A short-wave radio antenna is supported by two guy wires, 150 ft and 170 ft long. Each wire is attached to the top of the anten

na and anchored to the ground at two anchor points on opposite sides of the antenna. The shorter wire makes an angle of 65° with the ground. How far apart are the anchor points?

Physics
1 answer:
Nonamiya [84]3 years ago
4 0

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

You might be interested in
A woman experiences an electrical shock. The electrons making the shock come from the A woman experiences an electrical shock. T
Zinaida [17]

The electrons making the shock come from the women's body.

<h3>What is Electric shock ?</h3>

When a high voltage current flows through the body, electrical shock results. When someone unintentionally touches an electrical source, this typically occurs. Treatment for both internal and exterior burns may be necessary as part of the aftercare.

The nervous system may be impacted by a shock.

The tissue that makes up nerves presents extremely minimal resistance to the flow of an electric charge. Electric shocks that impact nerves can cause pain, tingling, numbness, weakness, or trouble moving a limb. These effects might disappear with time or remain for good.

How to Prevent from Electrical Shocks –

  • Keep the Appliances Away from Moisture and Water.
  • Never Connect or Disconnect Under Load.
  • Be Careful with Capacitors.
  • Use Insulated Tools.
  • Turn Off the Power.
  • Check for Improper or Faulty Wiring.
  • Fix Extension Cord Problems.

to learn more about electric shock go to - brainly.com/question/8822505

#SPJ4

7 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par
taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ =  5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 =   \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0
3 years ago
Last week we investigated the black hole at the center of our galaxy, with massMX= 8.5×1036kg. An object whose size is on the or
swat32

Answer:

d=2.38*10^{13}m

Explanation:

We know the mass of the hole, so we define as,

M_x= {8.5*10^{36}}kg

For which centripetal force is equal to gravitational force

\frac{mv^2}{r}=\frac{GMm}{d^2}

Angular velocity is equal to v/r,

w^2r=\frac{6.67*10^{-11}*8.5*10^{36}}{d^2}

As r=1 and w=1, we clear to d,

d^2=5.6695*10^{26}m

d=2.38*10^{13}m

6 0
4 years ago
We can model the human back as a pivoted rod?
3241004551 [841]

Answer:

So the answer is yes, we can the back be shaped like a spinning rod

spinal column that is approximated by a long and narrow rod,

Explanation:

The bone system of the body is very well modeled in physics, the back has a spinal column that is approximated by a long and narrow rod, this rod is fixed in the lower part to the coccyx and has a weight in the upper part (head), this rod has longitudinal vertical movement and twisting movement around the lower part of the bar.

So the answer is yes, we can the back be shaped like a spinning rod

7 0
3 years ago
Other questions:
  • PLEASE HELP!!!! Scientists launch a rocket, and they monitor its acceleration and the force exerted by its engines. As the rocke
    9·2 answers
  • The beat your doctor listens to through a sethoscope is the sound of the four values opening
    7·1 answer
  • A ______ is a network that has all connected devices located in the same physical location.
    8·1 answer
  • A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simu
    9·1 answer
  • The car's motion can be divided into three different stages: its motion before the driver realizes he's late, its motion after t
    12·1 answer
  • Now suppose the crate is lifted so rapidly that air resistance was significant during the raising. How much work was done by the
    13·1 answer
  • An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?
    11·1 answer
  • In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48
    5·1 answer
  • a student throws a coin vertically downward frok the top of a building. the coin leaves the throwers hand with a speed of 15.0m/
    5·1 answer
  • Why are carbon brushes used? How do these work?
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!