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Sedaia [141]
2 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]2 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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5 0
2 years ago
What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
lora16 [44]

Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or

<em>ρ</em> = <em>m</em> / <em>v</em>

Solving for <em>v</em> gives

<em>v</em> = <em>m</em> / <em>ρ</em>

So the given object has a volume of

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3 years ago
How many significant figures does the value 0.080 have? 1 2 3 4
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A box of volume V has a movable partition separating it into two compartments. The left compartment contains 3000 particles, the
storchak [24]

Answer:

a) V1 = 4V - V2/3 and V2 = 4V -  3V1

b) Δe = 4000V - 4000V2 + 9000V1

Explanation:

Let V represent volume of the box containing the two compartments

V1 represents compartment of the left compartment

V2 represents compartment of the right compartment

Momentum of the compartments before impact:

3000V1 + 1000V2

Momentum of the compartments after impact:

V(3000 + 1000) = 4000V

a) To obtain the volume of each compartment, that is, V1 and V2, we say:

Momentum before impact = Momentum after impact

3000V1 + 1000V2 = 4000V

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Also, V2 = 4000V - 3000V1/1000 = 4V - 3V1

b) Change in entropy,Δe = 4000V1 - 1000V2

By substituting the V1 and V2, we have:

4000(4V - V2)/3 - 1000(4V - 3V1)

16000V - 4000V2/3 - 4000V + 3000V1

16000V -  4000V2 - 12000V + 9000V1

∴ Δe = 4000V - 4000V2 + 9000V1

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