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nika2105 [10]
4 years ago
6

An airplane starting from airport a flies 399 km east, then 552 km at 58.5 ◦ west 3 of north, and then 890 km north to arrive fi

nally at airport
b. the next day, another plane flies directly from a to b in a straight line. in what direction should the pilot travel in this direct flight? use counterclockwise as the positive angular direction from due east, between the limits of −180◦ and 180◦ . assume there is no wind during these flights. answer in units of ◦ .
Physics
1 answer:
Lyrx [107]4 years ago
5 0

The approach here is to decompose all directions into North and East and then do simple Cartesian addition of the obtained coordinates.

S1 = 399 E

S2 = 552 Km 58.5 deg west of north ;

= 552 cos (58.5) N - 552 sin (58.5) E

= 288.41 N - 470.65 E

S3 = 890 N

total S = (890 + 288.41) N + (399 - 470.65) E

S = 1178.41 N - 71.65 E

D = \sqrt{N^2 + E^2}

D = 1180.58 Km

direction = atan(N/E) = -86.52 deg

means : 3.48 deg west of north

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Answer:

Assuming that you meant the final velocity of 50 m/s was reached in 10 s, the answer would be 5 m/s^2.

Explanation:

V_{f} = V_{i} + at

So we update that with the values that we have.

50 = 0 +a(10)

then simplify that using algebra to solve for a and we get 5 m/s^2

6 0
3 years ago
A thin-walled vessel of volume V contains N particles which slowly leak out of a small hole of area A. No particles enter the vo
LuckyWell [14K]

Answer:

    \frac{V}{2av}

Explanation:

From the question we are told that

Volume V

Contains N particles

Leaks from a small hole of area A

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Volume Flow Rate  V_r = A * v

Mathematically we find the  time taken to flow half way which is given by

          \frac{(V/2)}{A*v}

Therefore the  time taken is

           \frac{V}{2av}

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3 years ago
A body can have zero average velocity but not zero average speed. Why?​
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5 0
3 years ago
At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m
eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, t_w, is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, d_w, is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking t_w = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × t_w²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × t_w²

t_w = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = t_w ≈ 1.392 s

2) The horizontal distance, d_w, in meters from the edge of the pool to where the diver enters the water is given as follows;

d_w = L + v₀ × t_w = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0
3 years ago
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