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gavmur [86]
2 years ago
6

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

urface. The block is attached to a coil spring. The impact compresses the spring 15.0cm. Calibration of the spring shows that a force of 0.75N is required to compress the spring 0.25cm. Find the magnitude of the block\'s velocity just after the impact. What is the initial speed of the bullet?
Physics
1 answer:
mafiozo [28]2 years ago
6 0

Answer:

block velocity   v = 0.09186 = 9.18 10⁻² m/s  and speed bollet   v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

     p₀ = m v₀ + 0

After the crash

   p_{f} = (m + M) v

    p₀ = p_{f}

    m v₀ = (m + M) v                    (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

    Em₀ = K = ½ m v2

Final

    E m_{f}= Ke = ½ k x2

   Emo = E m_{f}

   ½ m v² = ½ k x²

   v² = k/m  x²

Let's look for the spring constant (k), with Hook's law

   F = -k x

   k = -F / x

   k = - 0.75 / -0.25

   k = 3 N / m

Let's calculate the speed

  v = √(k/m)   x

  v = √ (3/8.00)   0.15

  v = 0.09186 = 9.18 10⁻² m/s

This is the spped of  the  block  plus bullet rsystem right after the crash

We substitute calculate in equation  (1)

   m v₀ = (m + M) v

  v₀ = v (m + M) / m

  v₀ = 0.09186 (0.008 + 0.992) /0.008

  v₀ = 11.5 m / s

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It is mathematically expressed as:

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A bowling ball of mass 5.8 kg moves in a
evablogger [386]

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Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi
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Explanation:

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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that
Harrizon [31]

Answer:

The time taken is   t = 40007 sec  

Explanation:

From the question we are told that

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    The initial temperature of egg the T_e = 4.3^{o}C

     The temperature of the boiling water T_b = 100^oC

    The heat transfer coefficient is  H  = 800 W/m^2 \cdot K

    The  final temperature is T_e_f = 74^oC

     The  thermal  conductivity of water is k = 0.607 W/m^oC

     The diffusivity of the egg \alpha = 0.146 * 10^{-6} m^2 /s

Using one term approximation

We have the

            \frac{T_e_f - T_b}{T_e - T_b}  = Ae^{-\lambda ^2 \tau}

The radius is  r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m     Note that this radius is approximation to that of  a real egg

    Now we need to obtain the Biot number which help indicate the value of A  \ and \ \lambda to use in the above equation

     The Biot number is mathematically represented as

               Bi = \frac{H r}{k}

Substituting values  

               Bi = \frac{800 * 2.75 *10^{-2}}{0.607}

                    = 36.24

So for this value  which greater than 0.1 the  coefficient \lambda_1 \ and  \ A_1 is  

        \lambda = 3.06632

        A = 1.9942

Substituting this into equation 1 we have

          \frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-9.383 \tau}

           0.13624 =  e^{-9.383 \tau}

Taking natural log of both sides

           -1.993 =  -9.383\  \tau

          \tau =  0.2124

    The time required for the egg to be cooked is  mathematically represented as

          t = \frac{\tau r^2}{\alpha }

substituting value  is  

         = \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}

         t = 40007 sec  

8 0
3 years ago
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