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Sholpan [36]
3 years ago
7

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

tem is held in place. At time t=0 the blocks are released. The blocks move off in opposite directions with velocities v1 and v2. how much was the spring compressed?
Physics
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

The spring was compressed the following amount:

\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount \Delta x), and the total final energy is the addition of the kinetic energies of both masses:

E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2

E_i=E_f\\

\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

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Answer:

Please refer to the figure.

Explanation:

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5 0
3 years ago
I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
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Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR =  E- V  

                                                    IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

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7 0
3 years ago
A 4 kg bowling bowl is sitting on a table 1 meter off the ground. How much potential energy does it have?
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Answer:

\huge\boxed{\sf P.E. = 39.2\ Joules}

Explanation:

<u>Given Data:</u>

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

<u>Required:</u>

Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = mgh

<u>Solution:</u>

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
5 0
3 years ago
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