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vova2212 [387]
3 years ago
12

Ball 1, with a mass of 20kg is moving to the right at 20 m/s. At what velocity should ball 2, with a mass of 40kg, move so that

they both come to a standstill upon collision?
Physics
1 answer:
Ket [755]3 years ago
4 0
This is a conservation of momentum question. Initial momentum of the system is the momentum of ball 1 plus the momentum of ball 2. The final momentum of the system should be 0 since the balls stand still after the collision.

mv + mv = 0

mv = -mv

(20)(20) = -(40)Vi

400 = -40Vi

Vi = -10

So ball 2 should travel at 10m/s to the left
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Consider a metal single crystal oriented such that the normal to the slip plane and the slip directions are at angles of 43.1° a
Bond [772]

Answer:

resolve shear stress = 22 MPa

Explanation:

Given data

slip plane α = 43.1°

slip directions β = 47.9°

shear stress = 20.7 MPa (3,000 psi)

applied stress =45 mPa (6,500 psi)

to find out

what stress will be necessary

solution

we know that

resolve shear stress = aplied stress × cosα ×  cosβ

resolve shear stress = 45 × cos(43.1) ×  cos(47.9)

resolve shear stress = 22 MPa

we can say that here single cristal will be yield

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4 years ago
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Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu
Rus_ich [418]

Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is I  = 3.535 *10^{-6} \  W/m^2

Explanation:

From the question we are told that

    The temperature is  T = 2.81 \ K

Now  According to Stefan's law

        Power(P) =  \sigma  *  A  * T^4

Where  \sigma is the Stefan Boltzmann constant with value  \sigma  =  5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}

  Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

        I  =  \frac{P}{A}

=>      I  =  \frac{\sigma *  A  * T^4}{A}

=>      I  =  \sigma  *  T^4

substituting values

      I  = 5.67 *10^{-8}  *  (2.81)^4

       I  = 3.535 *10^{-6} \  W/m^2

       

4 0
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