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Licemer1 [7]
3 years ago
7

A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.

The captain orders the ship to travel 1.7 ✕ 106 kilometers, then turn 70° and travel an additional 2.7 ✕ 106 kilometers to reach the space station. If the captain had not ordered a course correction, what would have been the magnitude and direction of the path of the spaceship if it has traveled through the nebula?
Physics
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

Magnitude = 3.64 × 10^6  

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × 10^6    kilometers

turn = 70°

travel an additional = 2.7 × 10^6   kilometers

solution

we will consider here

Px = 1.7 × 10^6  

Py = 0

Qx =2.7 × 10^6  cos(70)

Qy= 2.7 × 10^6  sin(70)

so that

Hx = Px + Qx    ............1

Hx = 2.62 × 10^6  

and

Hy = Py + Qy      ..........2

Hy = 2.53 × 10^6  

so Magnitude = \sqrt{((2.62 \times 10^6  )^2+(2.53 \times 10^6)^2)}

Magnitude = 3.64 ×  

so direction  will be

tan စ = Hy ÷ Hx    ......................3

tan စ  = \frac{2.53}{2.62}

tan စ  = 0.9656

စ = 43.9°

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A) Work energy relation;
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A sprinter with a mass of 80 kg accelerates from 0 m/s to 9 m/s in 3 s. What is the runner's acceleration?
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3 years ago
A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the acceleration the car presents during this time?
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Answer:

1.02 m/s²

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0 m/s

Final velocity (v) = 6.6 m/s

Time (t) = 6.5 s

Acceleration (a) =.?

Acceleration can simply be defined as the change of velocity with time. Mathematically, it can be expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 6.6 m/s

Time (t) = 6.5 s

Acceleration (a) =.?

a = (v – u) / t

a = (6.6 – 0) / 6.5

a = 6.6 / 6.5

a = 1.02 m/s²

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2 years ago
A jet aircraft is traveling at 262 m/s in hor-
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Solution :

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Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

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Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$.

3 0
2 years ago
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