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Sedaia [141]
3 years ago
10

When a 2.40-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.92 cm.(a) Wh

at is the force constant of the spring
Physics
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

805.48N/m

Explanation:

According to Hookes law

F = Ke

F is the force = mg

F = 2.4×9.8 = 23.52N

e is the extension = 2.92cm = 0.0292m

Force constant K = F/e

K = 23.52/0.0292

K = 805.48N/m

Hence the force constant of the spring is 805.48N/m

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Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

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Studentka2010 [4]

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The athletes and sports persons possesses a good dynamic flexibility of their body as they carry our different types of body exercises.

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