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3 years ago

What is the average power output of an athlete who can life 9.0 * 10^2 kg 2.5 m in 2.0 s?

Physics

1 answer:

eduard3 years ago

3 0

Answer:

Average power output of athlete = 11025 Watts

Explanation:

Work done is defined as the product of force applied and the displacement perpendicular to the force.

Work = $Force\times Displacement$

Power is defined as work done per unit time.

Power = $\frac{Work done}{Time}$

Here the person lifts 900 kg.

Height = 2.5 m

Time interval = 2 seconds

Force = $weight \times gravity$

= $900 \times 9.8$

= 8820 N

Work done = $Force\times Displacement$

= $8820 \times 2.5$

= 22050 J

Power = $\frac {22050}{2}$

= 11025 Watt

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Science assessment help pls

DENIUS [597]

Answer:

Work Done = W

force = F

Distance = d

W = Fd

or W = F*d

W (in joules) = 3.5*4 = 14 Nm (or J)

1Nm = 1J

so newton meters and joules are the same

Power = Work (in joules) /time (in seconds)

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Answer:

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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit

sveticcg [70]

Answer:

a) I = 3.63 W / m² , b) I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

I = P / A = ½ ρ v (2π f $s_{max}$ )²

I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

cte = (½ ρ v 4π² s_{max}²)

I₀ = cte s² f₀²

I₀ = cte 10 6

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I = constant (2.20 10³) 2

I = cte 4.84 10⁶

let's find the relationship of the two quantities

I / Io = 4.84

I = 4.84 Io

I = 4.84 0.750

I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

I = cte (f s)²

I = constant (0.250 10³ 4)²

I = cte 1 10⁶

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I / Io = 1

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925=0(t)+1/2 (9.8)t^2

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