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Makovka662 [10]
4 years ago
13

Find the angle of depression from the top of a lighthouse 250 feet above water level to the water line of a ship 2.5 miles offsh

ore.
Physics
1 answer:
Rudik [331]4 years ago
4 0
Height of the lighthouse = 250 feet
Distance of the ship from the shore = 2.5 miles
We already know that
1 mile = 5280 feet
Then
2.5 miles = 5280 * 2.5
               = 13200 feet
Now, to find the angle of depression, we can use the formula
<span>θ = <span>tan<span>−1 </span></span>(13200/250)
</span>   = 1.50 degrees
I hope that this is the answer that you were looking for and the answer has come to your desired help. 
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A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?
balandron [24]

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s          Result

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3 years ago
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Answer:

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3 years ago
The more twists per foot in a pair of wires, the more resistant the pair will be to ____.
Murljashka [212]
The more twist per foot in a pair of wires, the more resistant the pair will be to cross talk. A cross talk in network planning and design is a disturbance produced by electromagnetic interference beside a circuit or a cable pair. A telecommunication signal interrupts a signal in an adjacent circuit and can source the signals to turn out to be confused and cross over each other.  
6 0
3 years ago
The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

<u>F = 3.86 x 10⁻⁶ N</u>

5 0
3 years ago
A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 0
3 years ago
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