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Virty [35]
3 years ago
12

A jogger leaves a point on a fitness trail running at a rate of 7 miles per hour. Five minutes later, a second jogger leaves fro

m the same location running at 8 miles per hour. How long will it take the second jogger to overtake the first?\
Physics
1 answer:
Liono4ka [1.6K]3 years ago
5 0

Answer:

Time taken is 0.5833 hours OR 35 minutes

Explanation:

In order to solve this equation we need to first find out how much distance the first jogger ran in the first five minutes. This is:

Distance of J1 = 7 * 5/60 = 0.5833 miles

The second jogger then leaves from the same location, and travels 0.5833 + (distance run by first jogger) to catch up.

We can write this as:

(Distance of J2) = 0.5833 + (Distance of J1)

Since distance = speed * time , we can write the above equation as follows:

(8 * time) = 0.5833 + (7 * time)

Solving for time we get:

Time = 0.5833 hours        (as velocity is in hours)

This is also equal to 35 minutes  

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A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and ampli
lys-0071 [83]

Answer:

(a) P_{avg} = 0.22 W

(b) P_{avg,2} = 0.056 W

Explanation:

given information:

the mass of piano wire, m = 3.00 g = 0.003 kg

tension, F = 25 N

length, l = 80 cm = 0.8 m

frequency, f = 120 Hz

amplitude, A = 1.6 mm = 0.0016 m

(a) the average power carried by the wave, P_{avg}

P_{avg} = \frac{1}{2}(√μF)ω²A²

where,

ω = 2πf = 2π120 = 754s^{-1}

μ = \frac{m}{l}

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  = 0.00375 kg/m

thus,

P_{avg} = \frac{1}{2}(√(0.00375)(25))(754)²(0.0016)²

P_{avg} = 0.22 W

(b) What happens to the average power if the wave amplitude is halved.

based on the equation above, we know that the average power is proportional to the square amplitude. therefore

\frac{P_{avg,1} }{P_{avg,2} } = \frac{A_{1} ^{2} }{A_{2} ^{2} } }

\frac{P_{avg,1} }{P_{avg,2} } = \frac{A_{1} ^{2} }{(0.5A_{1} )^{2} } }

P_{avg,2} = \frac{0.22}{4}

P_{avg,2} = 0.056 W

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3 years ago
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Explanation:

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Current I = ?

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Then, apply the formula for ohms law

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I = 24V / 60Ω

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6.3

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