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Evgen [1.6K]
3 years ago
14

How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 3.7 x 103 n/m?

Physics
2 answers:
Lubov Fominskaja [6]3 years ago
6 0
Hooke's law states that for a helical spring the extension is directly proportional to the force applied provided the elastic limit is not exceeded. 
Therefore; F= ke, where k is the spring constant, F is the force and e is the extension.
k = 2700 N/m and e = 3 cm or 0.03 M
therefore, F = 2700 × 0.03
                    = 81 N
Thus, the force required will be 81 N
AfilCa [17]3 years ago
5 0
From Hooke's Law 
We have F = ke where k is the spring constant and e is the extension.  
So e = 3cm = 0.03m and k is 3.7 * 10^3 
So it follows that F = 3.7 * 10^3 * 0.03 = 3.7 * 10^3 * 3 * 10^-2 
So F = 11.1 * 10^(3-2) = 11.1 * 10 = 111N
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Answer:

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One Newton is equivalent to<br> A. 1 kg/s2<br> B. 1 kg*m/s<br> C. 1 kg*m/s2<br> D. 1 kg/s
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2 years ago
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
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Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

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Acceleration = 414 m/s²

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Using newton's second law

F= ma

a=\dfrac{F}{m}

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7 0
3 years ago
Read 2 more answers
Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,  U_{f2} = 340.49 kJ/kg

(U_{fg} )_{2} =  2142.7 kJ/kg

Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

From the steam table,

S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

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enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

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Explanation:

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a) Using the principle of energy conservation,

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b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
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