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sineoko [7]
3 years ago
13

A soccer ball kicked on a level field has an initial vertical velocity component of 15.0 mps. assuming the ball lands at the sam

e height from which it was kicked was the total time of the ball in the air?

Physics
2 answers:
maxonik [38]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

jasenka [17]3 years ago
4 0

Answer:

3seconds

Explanation:

Time of flight of an object is the time taken by the object to spend in the air before landing. Mathematically it is represented as;

T = 2u/g

Where g is the acceleration due to gravity = 10m/s²

If the vertical velocity component (u) is given as 15m/s, the time of flight will be;

T = 2(15)/10

T = 30/10

T = 3seconds.

Therefore the total time by the ball in the air is 3seconds

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Infinite , is the correct answer for this question
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List two ways of recording or measuring absolute time
enyata [817]

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Stop watch

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3 years ago
While a block slides forward 1.35 m, a force pulls back at a 135 direction, doing -17.8 J of work. what is the magnitude of the
Mademuasel [1]
<h2>18.6467 N</h2>

Explanation:

       The Work done by any force is defined as the force applied times the displacement of point of application of force in the direction of force.

       This is better represented as a scalar product of Force vector and displacement vector.

       Work\textrm{ = }\vec{F}.\vec{s}

Here, the angle between force and displacement is 145^{o}.

       Work=|F||s|\cos\theta

       -17.8\textrm{ J = }|F|\times 1.35\textrm{ }m\times cos(135^{o})

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∴ Magnitude of force = 18.6467N

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3 years ago
Why is endurance training important to you?
Drupady [299]

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7 0
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Read 2 more answers
The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center. If
zvonat [6]

Answer:

The distance between their eye and the ceiling is 33.06 m.

Explanation:

Given that,

The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center, D = 4.35 mm

Diameter of the eye of pupil, d = 5.1 mm

Wavelength, \lambda=550\ nm

We need to find the distance between their eye and the ceiling. Using Rayleigh criteria, we get the distance as follows :

L=\dfrac{Dd}{1.22\lambda}\\\\L=\dfrac{4.35\times 10^{-3}\times 5.1\times 10^{-3}}{1.22\times 550\times 10^{-9}}\\\\L=33.06\ m

So, the distance between their eye and the ceiling is 33.06 m.

3 0
3 years ago
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