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Alchen [17]
4 years ago
12

She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ

ation deck is 1,353 feet high, how far apart are the boats?

Physics
1 answer:
Reika [66]4 years ago
3 0

Answer:

The boats are  934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i =  1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

<em><u>Step 1 : Finding the value of x(Refer the figure attached)</u></em>

We can use the tangent ratio to find the x value

tan(42^{\circ}) = \frac{1353}{x}

x = \frac{1353}{tan(42^{\circ}) }

x = 590.47 feet

<em><u>Step 2 : Finding the value of  z (Refer the figure attached)</u></em>

tan(29^{\circ}) = \frac{1353}{z }

z  = \frac{1353}{tan(29^{\circ})}

z = 1525.12  feet

<em><u>Step 3 : Finding the value of  y (Refer the figure attached</u></em>)

y =  z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

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A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th
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Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

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A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry
Sauron [17]

Answer:

a)  E = -4 10² N / C , b) x = 0.093 m, c)     a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

             F_{e} - T_{x} = m a

Axis y

            T_{y} - W = 0

Initially the system is in equilibrium, so zero acceleration

            Fe = T_{x}  

            T_{y} = W

Let us search with trigonometry the components of the tendency

            cos θ = T_{y} / T

            sin θ = T_{x}  / T

           T_{y} = cos θ

           T_{x}  = T sin θ

We replace

            q E = T sin θ

            mg = T cosθ

             

a) the electric force is

                F_{e} = q E

                E = F_{e} / q

                E = -0.032 / 80 10⁻⁶

                E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

                q E / mg = tan θ

                θ = tan⁻¹ qE / mg

Let's calculate

              θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

              θ = tan⁻¹ 0.3265

               θ = 18 ⁰

               sin 18 = x/0.30

               x =0.30 sin 18

               x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

           F_{e} = m aₓ

            aₓ = q E / m

           aₓ = 80 10⁻⁶ 4 10² / 0.01

           aₓ = 3.2 m / s²

Axis y

           W = m a_{y}

           a_{y} = g

           a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

             a = √ aₓ² + a_{y}²

             a = √ 3.2² + 9.8²

             a = 10.31 m / s²

The Angle meet him with trigonometry

               tan θ = a_{y} / aₓ

               θ = tan⁻¹ a_{y} / aₓ

               θ = tan⁻¹ (-9.8) / 3.2

               θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

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