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katovenus [111]
3 years ago
7

What is the net force at the equilibrium point? Derive an equation for the location of the equilibrium point based on the accele

ration due to gravity, the hanging mass, the spring constant and the position of the mass when the spring is unstretched.
Physics
1 answer:
Vitek1552 [10]3 years ago
4 0

To find a general equilibrium point for a spring based on the hook law, it is possible to start from the following premise:

Hook's law is given by:

F = k\Delta X

Where,

k= Spring Constant

\Delta X = Change in Length

F = Force

When there is a Mass we have two force acting at the System:

W= mg

Where W is the force product of the weigth. Then the force net can be defined as,

F_{net} = F+W

But we have a system in equilibrium, so

0 = K\Delta X -mg

We find the equilibrium for any location when

\Delta X = \frac{mg}{k}

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ExtremeBDS [4]

Atoms, molecules, electrons, photons, protons etc.

3 0
4 years ago
Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

5 0
3 years ago
Suppose you average 75 MPH over the first halfof a drive, and your average speed for the entiredrive is 60 MPH. What was your av
alexandr402 [8]

Answer:

x = 45 MPH

Explanation:

given,

Average speed of the first half = 75 MPH

Average speed of entire ride = 60 MPH

Average speed of the second half = ?

let the average speed of the second half = x MPH

now,

average of entire ride is given as 60 mph so,

 \dfrac{75+x}{2} = 60

 75+x= 2\times 60

       75 + x = 120

         x = 120 -75

        x = 45 MPH

hence, the average speed of the second half comes out to be 45 MPH.

5 0
3 years ago
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vladimir2022 [97]

Answer:

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Explanation:

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