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ch4aika [34]
3 years ago
11

When a plane flies with the​ wind, it can travel 1575 miles in 3.5 hours. when the plane flies in the opposite​ direction, again

st the​ wind, it takes 4.5 hours to fly the same distance. find the average velocity of the plane in still air and the average velocity of the wind?
Physics
1 answer:
Zepler [3.9K]3 years ago
3 0

When plane is flying along the wind then we can say

V_{plane} + v_{wind} = \frac{distance}{time}

V_{plane} + v_{wind} = \frac{1575}{3.5}

V_{plane} + v_{wind} = 450 mph

Now when its going against the wind the speed is given by

V_{plane} - v_{wind} = \frac{distance}{time}

V_{plane} - v_{wind} = \frac{1575}{4.5}

V_{plane} - v_{wind} = 350 mph

Now by the above two equations we will have

V_{plane} = 400 mph

v_{wind} = 50 mph

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Average speed of the runner is the rate at which the runner covers the total distance. Average speed of the runner in the race is given by,

Average speed = \frac{Total distance}{Total time}

Where

Total distance = Distance covered by the runner from initial to final position

Total time = time taken by the runner to cover entire distance

Instantaneous speed is the speed of the runner at the particular moment in the given time. Instantaneous speed is given by,

Instantaneous speed = \frac{dx}{dt}

x = position of the runner at time t

t = time taken to cover distance x

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6 0
3 years ago
A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds. 11-
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Answer:

a=2.4\ m/s^2

Explanation:

Given that,

Initial speed of a car, u = 13 m/s

Final speed of a car, v = 25 m/s

Time, t = 5 s

We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :

a=\dfrac{v-u}{t}\\\\a=\dfrac{25-13}{5}\\\\=2.4\ m/s^2

So, the acceleration of the car is 2.4\ m/s^2.

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3 years ago
All the bulb in the series circuit are on or off together?
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When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of
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Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

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3 years ago
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azamat

Answer:

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Explanation:

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