A pilot in an airplane flying at 25,000 ft sees two towns directly ahead of her in a straight line. The angles of the depression
to the towns are 25o and 50o , respectively. To the nearest foot, how far apart are the towns?
2 answers:
Answer:
The towns are 32,635 ft apart.
Explanation:
From the image drawn below:
AB = x ft
BC = a ft
AC = (x + a) ft
Considering triangle PCB,
tan 50° = 25000 / a
Or,
a = 25000 / tan 50°
Since tan x = 1/ cot x
a = 25000×cot 50°------------------------------------1
Considering triangle PCA,
tan 25° = 25000 / (a + x)
Or,
a + x = 25000 / tan 25°
Since tan x = 1/ cot x
a + x = 25000×cot 25° -------------------------------2
Thus, finding x from equation 1 and 2, we get:
x = 25000 (cot 25° - cot 50°)
Using cot 25° = 2.1445 and cot 50° = 0.8394, we get:
<u>x ≈ 32,635 ft</u>
<u>Thus, the distance between two towns is 32,635 ft.</u>
Answer:
32635.17 ft
Explanation:
In the diagram, AB = 25000 ft
Let C and D be the town, where CD = d (Distance between two towns)
By triangle, ABC
tan 50 = AB / AC
AC = 25000 / tan 50 = 20977.5 ft
By triangle, ABD
tan 25 = AB/AD
AD = 25000 / tan 25 = 53612.67 ft
So, the distance between two towns
d = AD - AC = 53612.67 - 20977.5 = 32635.17 ft
Thus, the distance between two towns is 32635.17 ft.
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• the normal force with magnitude <em>n</em>
• the car's weight with magnitude <em>w</em>
and the net force points toward the center of the circle made by the turn, with centripetal acceleration
<em>a</em> = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²
Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,
∑ <em>F</em> (⟂) = <em>N</em> + <em>W</em> (⟂) = <em>m</em> <em>a</em> (⟂)
and
∑ <em>F</em> (//) = <em>W</em> (//) = <em>m a</em> (//)
Let the direction of <em>N</em> be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle <em>θ</em> with the banked curve, and <em>W</em> makes the same angle with the negative perpendicular axis, so that the equations above reduce to
<em>N</em> - <em>m g</em> cos(<em>θ</em>) = <em>m</em> <em>a</em> sin(<em>θ</em>)
and
<em>m g</em> sin(<em>θ</em>) = <em>m a</em> cos(<em>θ</em>)
The second equation is all we need at this point to find the ideal <em>θ</em>. The mass <em>m</em> cancels out, and we can solve for <em>θ</em> to get
tan(<em>θ</em>) = <em>a</em>/<em>g</em> ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615
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Answer:
Explanation:
We are asked to find the force applied to the volleyball. According to Newton's Second Law of Motion, force is the product of mass and acceleration.
We are given the mass, but we must calculate the acceleration.
1. Acceleration
Acceleration is the rate of change of velocity with respect to change. It is the change in velocity over the time.
The volleyball starts at rest or 0 meters per second and reaches a velocity of 22 meters per second in 0.2 seconds.
= 22 m/s
= 0 m/s
- t= 0.2 s
Substitute the values into the formula.
Solve the numerator.
Divide.
2. Force
Now we know the mass and acceleration, so we can calculate the force.
The mass is 0.5 kilograms and the acceleration is 110 meters per second squared.
- m= 0.5 kg
- a=110 m/s²
Substitute the values into the formula.
Multiply.
1 kilogram meter per second squared is equal to 1 Newton, so our answer of 55 kg*m/s² is equal to 55 Newtons.
<u>55 Newtons</u> of force are applied to the volleyball.