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FinnZ [79.3K]
4 years ago
7

What will a spring scale read for the weight of a 75.0-kg woman in an elevator that moves upward with constant speed of 5.8 m/s

?
Physics
1 answer:
Sholpan [36]4 years ago
7 0
<span>On the scale the only external forces are the man's weight acting downwards and the normal force which the scale exerts back to support his weight. So F = Ma = mg + Fs The normal force Fs (which is actually the reading on the scale) = Ma + Mg But a = 0 So Fs = Mg which is just his weight. Fs = 75 * 9.8 = 735N</span>
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Explanation:

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y_2 = (6 mm) sin(2x - t)

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y_1 = (3 mm) sin(x - 3t) ,  y_4 = (2 mm) sin(x - 2t) ,y_2 = (6 mm) sin(2x - t) ,  y_3 = (1 mm) sin(4x - t)

b )

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3 years ago
Left arm has area 10.0 cm2, right has area 5.00 cm2. 100 g of water isadded to right side. a) Determine the length of the water
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Answer:

0.49  cm

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h_2=\frac{hA_1}{A_2}

13.6\times 9.8(h+h(\frac{A_1}{A_2}))=1\times 9.8\times 20

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h=\frac{20A_2}{13.6(A_2+A_1)}

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