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Makovka662 [10]
3 years ago
13

The velocity of a plane that travels 450.0 km due north in 3.500 hours is

Physics
2 answers:
dalvyx [7]3 years ago
6 0

Answer : The velocity of a plane that travels 450.0 km due north in 3.500 hours is, 35.71 m/s

Solution : Given,

Distance = 450 km = 450000 meter     (1 km = 1000 m)

Time = 3.5 hours = 12600 second        (1 hr = 3600 s)

Velocity : It is defined as the displacement divided by the time. It is vector quantity. That means both magnitude and the direction are needed to define the velocity. The S.I unit of velocity is meter per second (m/s).

Formula used :

Velocity=\frac{Displacement}{Time}

Now put all the given values in this formula, we get

Velocity=\frac{450000m}{12600s}=35.71m/s

Therefore, the velocity of a plane that travels 450.0 km due north in 3.500 hours is, 35.71 m/s

ddd [48]3 years ago
4 0
I believe it's <span> 128.6 km/h due north. 
</span>
I hope this helps!
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someone help pls. Two students, Mia and Peter, leave school to meet at the local coffee shop. Peter decides to jog to the coffee
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Answer:

1) The distance further it takes Peter to arrive at the Coffee shop than Mia is 1.24 km

2) Mia's average speed is 6.00 km/hour

Peter's average speed is 8.48 km/hour

4) Mia's average velocity = Peter's average velocity = 6.00 km/hour

Explanation:

The given information from the diagram are;

The distance Peter jogs from school to the flower shop = 2.00 km

The distance Peter jogs from the Flower shop to the Coffee shop = 2.24 km.

The distance Mia walks from school directly to the Coffee shop = 3.00 km

The time it takes both Peter and Mia to arrive at the coffee shop = 30 minutes = 0.5 hour

1) The total distance Peter travels from school to the Coffee shop = 2.00 km + 2.24 km = 4.24 km

The distance Mia travels from school to the Coffee shop = 3.00 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 4.24 km - 3.00 km = 1.24 km

The distance further it takes Peter to arrive at the Coffee shop than Mia = 1.24 km

2) Average \ speed = \dfrac{Total \ distance \ traveled}{Total \ time \ taken \  in \ the \ journey}

Therefore, \ Mia's \ average \ speed = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average speed = 6.00 km/hour

Peter's \ average \ speed = \dfrac{4.24 \ km}{0.5 \ hour}= 8.48 \ km/hour

Peter's average speed = 8.48 km/hour

4) Average \ velocicty = \dfrac{Displacement }{Time  \ taken}

The displacement from the School to the Coffee shop is 3.00 km for both Mia and Peter

The time it takes both Peter and Mia to arrive at the Coffee shop from the school is 30 minutes = 0.5 hour

Therefore, \ Mia's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Mia's average velocity = 6.00 km/hour

Peter's \ average \ velocity = \dfrac{3.00 \ km}{0.5 \ hour}= 6.00 \ km/hour

Therefore, Peter's average velocity is also = 6.00 km/hour

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Oksanka [162]
Air resistance is the answer
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A net force of 3 N accelerates a mass of 3 kg at the rate of 1 m/s2. The acceleration of a mass of
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Answer:

,mkjh ,bkl m,

Explanation:

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Which scientist is often called the “father of the atomic bomb” because of his work as the head of the manhattan project?
kirza4 [7]

Answer:

J. Robert Oppenheimer

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Help meh in this question​
V125BC [204]

  • Radius=r=R_o/2
  • angular velocity=w=v_o
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We know

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

For Rotational motion

\boxed{\sf v=r\omega}.

Putting value

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{2}m(r\omega)^2

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{2}mr^2\omega^2

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{2}m\left(\dfrac{R_o}{2}\right)^2v_o^2

\\ \qquad\quad\sf{:}\dashrightarrow K.E=\dfrac{1}{4}mv_o^2

7 0
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