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lidiya [134]
2 years ago
8

From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o

f the boat is θ = 4 ◦ . How far (in feet) is the boat from the shoreline? Answer in units of feet.

Physics
1 answer:
maksim [4K]2 years ago
4 0

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

tangent( O ) = \frac{opposite}{adjacent}

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

tangent( 4) = \frac{53 feets}{DB}

DB=  \frac{53 feets}{tangent( 4)}

DB=  757,93 feets

This its equal to the distance from the boat to the shoreline.

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A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"
Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0
2 years ago
Joey is riding in an elevator which is accelerating upwards at 2.0 m/s2. The elevator weighs 300.0 kg, and Joey weighs 60.0 kg.
vodka [1.7K]

4200 N  is the tension in the cable that pulls the elevator upwards.

The correct option is A.

<h3>What does tension ?</h3>

Tension is the force that is sent through a rope, thread, or wire whenever two opposing forces pull on it. Along the whole length of the wire, the tensile stress pulls equally on all objects at the ends. Every physical object that comes into contact with that other one exerts force on it.

<h3>Briefing:</h3>

We employ the following formula to determine the cable's tension.

Formula:

T = mg+ma............ Equation 1

Where:

T is the cable's tension.

M = Mass of the elevator and the Joey

Accelerating with a

g = Gravitational acceleration

Considering the query,

Given:

m = (300+60) = 360 kg

a = 2 m/s²

g = 9.8 m/s²

Substitute these values into equation 2

T = (360×9.8)+(360×2)

T = 3528+720

T = 4248 N

T ≈ 4200 to the nearest hundred.

To know more about Tension visit:

brainly.com/question/14177858

#SPJ1

7 0
1 year ago
A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black
Wewaii [24]
Red light reflects off the card into your eyes and you see the red card as red. The light will just make the card brighter. So A
4 0
2 years ago
If a car traveled 60 m in 30 s, what is the velocity?*
klemol [59]

Answer:

6.56168

Explanation:

8 0
2 years ago
Spaceship A is 10 meters long and approaching you from the south at a speed of.7c while spaceship B, which is also 10 meters lon
Valentin [98]

Answer:

a) 0.94 C

b) 7.14 m

c) 3.11 m

d) 1.40 s

e) 2.93 s

Explanation:

First we need to set up a coordinate system. This will have the positive X axis pointing north. So spaceship A has positive speed, and spaceship B has negative speed.

The Lorentz transformation for speed is:

u' = \frac{u - v}{1 - \frac{u*v}{c^2}}

u: speed of spaceship A as observed by you

v: speed of spaceship B as observed by you

In the case of the speed of spaceship A as observed by spaceship B:

u' = \frac{0.7c - (-0.7c)}{1 - \frac{0.7c*(-0.7c)}{c^2}} = 0.94c

The transform for lengths is:

L = L0 * \sqrt{1 - \frac{v^2}{c^2}}

For the case of spaceship A as observed by you:

L = 10 m * \sqrt{1 - \frac{(0.7c)^2}{c^2}} = 7.14 m

For the case of spaceship A as observed by spaceship B:

L = 10 m * \sqrt{1 - \frac{(0.94c)^2}{c^2}} = 3.12 m

The time dilation equation is:

T = \frac{T0}{\sqrt{1-\frac{v^2}{c^2}}}

For the case of the event as observed by you:

\frac{1 s}{\sqrt{1-\frac{(0.7c)^2}{c^2}}} = 1.40 s

For the case of the event as observed by spaceship B:

\frac{1 s}{\sqrt{1-\frac{(0.94c)^2}{c^2}}} = 2.93 s

3 0
3 years ago
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