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BabaBlast [244]
2 years ago
11

1. A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ba

ll, it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?
Physics
1 answer:
MariettaO [177]2 years ago
8 0
Momentum = (mass) x (velocity)

Original momentum before the hit =

                   (0.16 kg) x (38 m/s) this way <==

               =             6.08 kg-m/s  this way <==

Momentum after the hit =

                   (0.16) x (44 m/s) that way  ==>

               =           7.04 kg-m/s  that way ==>

Change in momentum = (6.08 + 7.04) =  13.12 kg-m/s  that way ==> .
-----------------------------------------------

Change in momentum = impulse.

                                   Impulse = (force) x (time the force lasted)

                          13.12 kg-m/s  = (force) x (0.002 sec)

  (13.12 kg-m/s) / (0.002 sec)  =  Force

             6,560 kg-m/s² = 6,560 Newtons  =  Force   

                            ( about 1,475 pounds  ! ! ! )
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5 0
3 years ago
The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current
Komok [63]

Answer:

I=0.047A

Explanation:

Let's use Ohm's law:

V=IR  

or

I=\frac{V}{R}   (1)

Where:

V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

R=\rho*\frac{l}{A}    (2)

Where:

R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}

Keep in mind that the electrical resistivity of the gold is a known constant which is \rho_g_o_l_d=2.35*10^{-8} and the cross sectional area of the conductor is calculated as:

A=\pi *(r^{2})=\pi  *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} }  =14.96056465\Omega

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

I=\frac{0.7}{14.96056465}\approx0.047A

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