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Karo-lina-s [1.5K]
3 years ago
11

3. The center of mass (or center of gravity) of a two-particle system is at the origin. One particle

Physics
1 answer:
nika2105 [10]3 years ago
8 0

Answer:

B) (-2.0 m, 0.0 m)

Explanation:

Given:

Mass of particle 1 is, m_1=2.0\ kg

Mass of particle 2 is, m_2=3.0\ kg

Position of center of mass is, (x_{cm},y_{cm})=(0,0)

Position of particle 1 is, (x_1,y_1)=(3.0\ m,0.0\ m)

Position of particle 2 is, (x_2,y_2)=(?\ m,?\ m)

We know that, the x-coordinate of center of mass of two particles is given as:

x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

Plug in the values given.

0=\frac{(2.0\times 3)+(3.0\times x_2)}{2.0+3.0}\\\\0=6+3x_2\\\\3x_2=-6\\\\x_2=\frac{-6}{3}=-2.0\ m

We know that, the y-coordinate of center of mass of two particles is given as:

y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}

Plug in the values given.

0=\frac{(2.0\times 0)+(3.0\times y_2)}{2.0+3.0}\\\\0=0+3y_2\\\\3y_2=0\\\\y_2=\frac{0}{3}=0.0\ m

Therefore, the position of particle 2 of mass 3.0 kg is  (-2.0 m, 0.0 m).

So, option (B) is correct.

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Answer:

\tau=3.3*10^{-6}s

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

C*\frac{dV}{dt}+\frac{V}{R}=0

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

\frac{dV}{V}=-\frac{1}{RC}dt

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt

Evaluating the integrals:

ln(\frac{V}{v})=e^{\frac{-t}{RC} }

natural logarithm to both sides in order to isolate V:

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3 0
3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
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Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

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<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

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We have seen and recorded many such eclipses on Jupiter. These are from the perspective of Earth. When the moons of Jupiter comes exactly between the Sun and Jupiter the shadow of moon will fall on Jupiter. The places where the shadow falls, one will see a solar eclipse.

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