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strojnjashka [21]
2 years ago
5

The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports

that the wind is coming from the west at 58 miles per hour. in order to travel in a straight line, the ring wraiths decide to head northwest. at what speed should they fly (omit units)?
Physics
2 answers:
Katarina [22]2 years ago
5 0
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.

c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02 
 
Alternative solution:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Therefore, they have to fly 82.02 mph
agasfer [191]2 years ago
5 0

Answer:

V = 82.02 mph

Explanation:

Speed of the wind, v = 58 mph

We need to find the speed with which it fly. It will form a right angled triangle. It can be calculated as :

V=\sqrt{58^2+58^2}

V = 82.02 mph

So, the speed of the wind is 82.02 mph. Hence, this is the required solution.

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The colored lines in the figure represent paths taken by different people walking around in a city. Assume that each city block
jarptica [38.1K]

Answer: 592.37m

Explanation:

Person D is the blue line.

The total displacement is equal to the difference between the final position and the initial position, if the initial position is (0,0) we have that he first goes down two blocks, then right 6 blocks. then up 4 blocks, then left 1 block.

Now i will considerate that the positive x-axis is to the right and the positive y-axis is upwards.

Then the new position will be, if B is a block:

P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)

And we know that B = 110m

P = (550m, 220m)

Now, then the displacement will be equal to the magnitude of our vector, (because the difference between P and the initial position is equal to P, as the initial position is (0,0))  this is:

P = √(550^2 + 220^2) = 592.37m

4 0
2 years ago
As the drawing illustrates, a siren can be made by blowing a jet of air through 20 equally spaced holes in a rotating disk. The
Aneli [31]

Answer:

ω = 630.2663 = 630[rad/s]

Explanation:

Solution:

- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.

                                  1rev(20 hole) -> 20(cycle)/rev  

                                        2006.2(cycle) -> f ?  

                              f = 2006.2/20 = 100.31rev at second  

- The relation between angular frequency and angular speed is given by:

                                 ω = 2πf

                                 ω = 2*3.14*100.31

                                 ω = 630.2663 = 630[rad/s]

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8 0
2 years ago
A sharp edged orifice with a 50 mm diameter opening in the vertical side of a large tank discharges under a head of 5m. If the c
Sever21 [200]

Answer:

0.24

Explanation:

See attached file

3 0
2 years ago
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
2 years ago
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