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soldier1979 [14.2K]
3 years ago
7

What is the acceleration of a full bottle of water dropped fron a desk​

Physics
1 answer:
Elodia [21]3 years ago
8 0

The acceleration of the bottle is g=9.8 m/s^2 downward

Explanation:

When an object is in free fall (=only the force of gravity is acting on it), its acceleration is equal to the acceleration of gravity.

In fact, the only force acting on the object is the force of gravity, whose magnitude is:

F=mg

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

According to Newton's second law, the net force on the object is equal to the product between its mass (m) and its acceleration (a), so

F=ma

By combining the two equations, we get

mg=ma\\a = g

Therefore, the acceleration of any object in free fall (and so, also of the bottle in the problem) is g=9.8 m/s^2.

Learn more about weight and forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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c, 280,000J

Explanation:

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How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00
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Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

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It is known that the dielectric strength of air is as follows.

               E = 3 \times 10^{6} V/m

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = \frac{\epsilon_{o} A}{d} \times E \times d

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                   = 8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}

                   = 1.062 \times 10^{-8} C

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Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th
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Complete Question

Diagram for this  shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is I{xx}= 0.119 kg \cdot m^2

Explanation:

From the question we are told that

        The density `of wooden handle is  \rho_w = 860 kg/m^3

        The density `of soft-metal head  is \rho_s =8000kg/m^3

Generally the mass of the wooden can be mathematically obtained with this formula

          m_w = \rho_w A_w l_w

Where A_w is mass of wooden handle which is  mathematically obtain with the formula

             A_w = \frac{\pi}{4} d^2_w

Where d_w is the diameter  of the wooden handle which from the diagram is

       27mm = \frac{27}{1000} = 0.027m

So  A_w = \frac{\pi}{4} * 0.027^2

      l_w is the length of the the wooden handle which is given in the diagram as   l_w = 315mm = \frac{315}{1000} = 0.315m

Substituting these value into the formula for mass

      m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315

            = 0.155kg

Generally the mass of the soft-metal head can be mathematically obtained with this formula

           m_s = \rho_s A_s l_s

Where A_s is mass of soft-metal head which is  mathematically obtain with the formula

            A_s = \frac{\pi}{4} d^2_s

Where d_s is the diameter  of the soft-metal head which from the diagram is            

       36mm = \frac{36}{1000} = 0.036m

So  A_s = \frac{\pi}{4} * 0.036^2

 l_s is the length of the the soft-metal head which is given in the diagram

     as   l_s = 90mm = \frac{90}{1000} = 0.090m

Substituting these value into the formula for mass  

                  m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090

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Substituting values

                   (I_{xx})_w  =    [\frac{1}{3}*0.155 + 0.315^2 ]

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Substituting values

                 b = 0.315 + \frac{0.036}{2}

                    = 0.336m

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

                            (I_{xx})_s = [\frac{1}{12} *0.733*  0.090^2 + 0.336^2]

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         I_{xx} = (I_{xx})_w + (I_{xx})_s

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Substituting values

        I_{xx} = 5.12*10^{-3} +0.113

               I{xx}= 0.119 kg \cdot m^2

             

             

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