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4 years ago

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Physics

2 answers:

Otrada [13]4 years ago

7 0

The energy stored in a spring is

E = 1/2 (k x²)

' k ' is the spring constant . . . the number of newtons of force it takes to stretch or compress the spring by one meter from its relaxed length

' x ' is the distance the spring is stretched or compressed

For this problem, we know ' E ' and ' x ', and we need to find ' k '.

E = 1/2 (k x²)

1,000 = 1/2 (k) (0.025²)

1,000 = 1/2 (k) (0.000625)

Multiply each side by 2 :

2,000 = (k) (0.000625)

Divide each side by 0.000625:

(2,000) / (0.000625) = k

k = 3,200,000

The spring constant is 3.2 x 10⁶ Newtons per meter .

This is a very stiff spring ! My calculator says that if you want to stretch it just 1 inch, you have to pull it with a force of 18,285 pounds ! ! If such a spring exists, it might be used as part of the suspension for a tank or a concrete truck.

ziro4ka [17]4 years ago

6 0

Answer:

Explanation:

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acontainer is filled whith mercury to alevel of 10m whit water to alevel of 8m and whit oil to alevel of 5m the densities oil ,w

Alborosie

Answer:

1450.4 KN $m^{2}$

Explanation:

Pressure = ρhg

where: ρ is the density of the liquid, h is the height and g the force of gravity.

Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury

So that,

i. Pressure of oil = ρhg

(ρ = 0.8 g/cm³ = 800 kg/m³)

= 800 x 5 x 9.8

= 39200

Pressure of oil = 39200 N $m^{2}$

ii. Pressure of water = ρhg

(ρ = 1 g/cm³ = 1000 kg/m³)

= 1000 x 8 x 9.8

= 78400

Pressure of water = 78400 N $m^{2}$

ii. Pressure of mercury = ρhg

(ρ = 13.6 g/cm³ = 13600 kg/m³)

= 13600 x 10 x 9.8

= 1332800

Pressure of mercury = 1332800 N $m^{2}$

So that,

Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800

= 1450400

= 1450.4 KN $m^{2}$

Total pressure exerted by the liquids at the base is 1450.4 KN $m^{2}$ .

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