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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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RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s
Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, R=50\ K\Omega=5\times 10^4\ \Omega

Capacitance, C=2.1\ \mu F=2.1\times 10^{-6}\ F

We need to find the expected time constant for this RC circuit. It can be calculated as :

\tau=R\times C

\tau=5\times 10^4\times 2.1\times 10^{-6}

\tau=0.105\ s

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

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List 5 reason why water is not thermometric liquid​
kolezko [41]

Answer:

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4 years ago
You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1
drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

  • <em>Mass of the pendulum, = M </em>
  • <em>Length of the pendulum, = L</em>
  • <em>Initial angular speed, </em>\omega _i<em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

I_i = \frac{1}{3} ML^2

The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

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3 years ago
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