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3 years ago

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A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 1.9 m/s^2. The acceleration of gravity is 9.8

m/s^2. What is the magnitude of the force of the elevator floor on the person? Answer in kN and round 4 decimal places.

1 answer:

Keith_Richards [23]3 years ago

3 0

Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

$F = m\cdot(g-a)$

We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

$900 N = m\cdot g \implies m = \frac{900N}{9.8 \frac{m}{s^2}}$

So

$F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN$

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$D = (v_1 * t_1) + (v_2 * t_2 ) + (v_3 * t_3)$

substituting values

$D = (7.2 * 1320) + (5.1 * 2160) + (13 * 480)$

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Generally the average velocity is mathematically represented as

$v_{avg} = \frac{D}{t_{total}}$

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$t_{total} = 1320 + 2160 + 480$

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The average velocity is

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The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

Mademuasel [1]

Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

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$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

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$100dB=10 log(\dfrac{r_02}{(5m)^2})$

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$10^{10}= \dfrac{r_0^2}{25}$

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Now equation (2) becomes

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$85 = 10log(\dfrac{25*10^{10}}{r^2})$

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