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ruslelena [56]
2 years ago
8

An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

tially. if the mass of the second cart is m2=0.43kg, how much kinetic energy is lost as a result of the collision?
Physics
1 answer:
arsen [322]2 years ago
7 0
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
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The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2
masha68 [24]

Answer:

the normal stress induced by  the concrete post \sigma_c = 67.26 psi

the normal stress induced by the steel \sigma_s = - 1795.84 psi

Explanation:

Given that:

Modulus for elasticity for concrete post E_c = 3.6 *10^6 psi

Thermal coefficient for concrete post  \alpha _c = 5.5 *10^{-6}/^0F

Modulus for elasticity of steel bar E_s = 29*10^6psi

Thermal coefficient of steel bar \alpha _2 = 6.5*10^{-6}/^0F

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod A_s = 6(\frac{ \pi}{4} )(d_s)^2

= 6(\frac{ \pi}{4} )(\frac{7}{8} )^2

= 3.61 in²

Area of concrete parts A_c = (10)(10) - A_s

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= [\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)

= [\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)

= [(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}

= 1.234*10^{-8}P = 8.0*10^{-5}

P = \frac {8.0*10^{-5}}{1.234*10^{-8}}

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

\sigma_c =\frac{P}{A_c}

\sigma_c = \frac{6482.98}{96.39}

\sigma_c = 67.26 psi

Thus, the normal stress induced by  the concrete post \sigma_c = 67.26 psi

Also; the normal stress in the steel bars  induced as a result of temperature rise is as follows:

\sigma_s = \frac{-P}{A_s}

\sigma_s =\frac{-6482.98}{3.61}

\sigma_s = - 1795.84 psi

Thus, the normal stress induced by the steel \sigma_s = - 1795.84 psi

6 0
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8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti
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Answer:

EMF induced in the loop is 9.4 V

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As we know that initial magnetic flux of the loop is given as

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\phi_1 = (0.60)(\pi (0.05)^2)

\phi_1 = 4.7 \times 10^{-3} Wb

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Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

EMF = \frac{\Delta \phi}{\Delta t}

so we will have

EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}

EMF = 9.4 V

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