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hichkok12 [17]
3 years ago
11

Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i

nitially traveling away from Aat 2.0m/s. After the collision the center of mass of the two cart system has a speed of:____________.A. zeroB. 0.33m/sC. 2.3m/sD. 2.5m/sE. 5.0m/s

Physics
1 answer:
inysia [295]3 years ago
3 0

Answer:

\large \boxed{\text{C. 2.3 m/s}}

Explanation:

Data:

m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\

Calculation:

This is a perfectly inelastic collision.  The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}}  + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}

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T = 1.108\,s

Explanation:

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The length of the leg is approximately the height of the person:

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The period is:

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3 years ago
An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it
steposvetlana [31]

Answer:

μs = 0.36

Explanation:

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  • This acceleration is produced by the centripetal force.
  • Now, this force is not a different type of force, is the net force acting on the riders in this direction.
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  • In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
  • When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

       F_{frmax} = \mu_{s}* F_{n}  = m * g  (1)

  • where μs= coefficient of static friction (our unknown)
  • As  we have already said Fn = Fc.
  • The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

      F_{n} = m* \omega^{2} * r  (2)

  • Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

       \mu_{s} = \frac{g}{\omega^{2} r}  (3)

  • Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

       \omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)

  • Replacing g, ω and r in (3):
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2 years ago
Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magni
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3 0
3 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0
3 years ago
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