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2 years ago

A soccer player kicks a ball with a speed of 30 m/s at an angle of 10. How long does the ball stay in the air?

1 answer:

4 0

"10" is not a very satisfying description of an angle.
I'll assume it's 10 degrees, and that it's the angle of the initial trajectory
above the horizontal.

The vertical component of the ball's speed is 30 sin(10°) = 5.209 meters per second.

The acceleration of gravity is 9.8 meters per second² downward.

The ball continues to rise for 9.8/5.209 = 1.881 seconds after the kick.
At that time, it reaches the peak of its arc, and if nobody else has contacted it yet,
it begins to fall, and takes the same length of time to return to the ground whence
it was kicked.

So the time between the kick and return to the ground is (2 x 1.881) = <u>3.763 seconds</u>.

Note: As is customary and necessary at the high school level,
we assume no air resistance.

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3 years ago

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2 years ago

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Answer:

Δ $R_{e}$ = 84 Ω, $R_{e}$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_{e}$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_{e}$ = 1 / R₁ + 1 / R₂

1 / $R_{e}$ = 1/500 + 1/2000 = 0.0025

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Let's look for the error (uncertainly) of Re

$R_{e}$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_{e}$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_{e}$ / $R_{e}$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_{e}$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_{e}$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_{e}$ = 0.21 400

Δ $R_{e}$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_{e}$ = (40 ± 8) 10¹ Ω

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2 years ago

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