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dimulka [17.4K]
2 years ago
10

What is deceleration?????Have a nice day​

Physics
2 answers:
const2013 [10]2 years ago
6 0

Answer:

\deceleration)

Explanation:

<h2>DEÇELERATIØÑ ALWAYS REFERs TØ ACCELERATION IN THE DIRECTION ØPPOSITE TO THE DIRECTION OF VILØÇÏTY <em><u>.</u></em></h2>

<h3><em><u>DEÇELERATIØÑ</u></em><em><u> </u></em><em><u>ALWAYS</u></em><em><u> </u></em><em><u>reduces</u></em><em><u> </u></em><em><u>speed</u></em><em><u> </u></em><em><u>.</u></em></h3>

<em><u>✌️</u></em><em><u>ur</u></em><em><u> answer</u></em><em><u> hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> you</u></em><em><u> ✌️</u></em>

Rufina [12.5K]2 years ago
3 0
Deceleration—the ability to slow down and control force production—is often ignored during training; but deceleration technique is critical for most sports. Speed is often the factor that separates the elite from the average athlete. Credit source is stack.com have a nice day! I told you it was from the internet in case you couldn’t use the internet Hope this helped :)!
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2 years ago
baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

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The magnitude of the electric force between two protons is 2.30 10-26 n. how far apart are they?
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The electric force between two charge objects is calculated through the Coulomb's law.
                               F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
                           2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m. 
7 0
2 years ago
Read 2 more answers
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