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BabaBlast [244]
3 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]3 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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Explanation

1. <u>Data</u>:

a) ω = constant

b) Abbie: r₁ = 1 m

c) Zak: r₂ = 2 m

d) Ac₁ = ? Ac₂

2. <u>Formulae</u>

  • Ac = ω² r

3. <u>Solution</u>:

a) Abbie:

  • Ac₁ = ω² r₁  =  ω² (1m)

b) Zack:

  • Ac₂ = ω² r₂  = ω² (2m)

c) Divide Ac₁ / Ac₂

  • Ac₁ / Ac₂ =  ω² (1m) / [ω² (2m) ] = 1/2

⇒      Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂

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Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
3 years ago
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