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stellarik [79]
3 years ago
10

The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration ph

ase lasts only 50 ms, during which the ball's speed increases from 0 to about 90 mph, or 40 m/s.
What is the force of the pitcher's hand on the 0.145 kg ball during this acceleration phase? Express your answer with the appropriate units.
Physics
1 answer:
SashulF [63]3 years ago
4 0

Answer:

116 N.

Explanation:

Given,

mass of the ball, m = 0.145 Kg

initial speed = 0 m/s

Final speed = 40 m/s

time = 50 ms = 0.05 s

Force is equal to change in momentum per unit time

F = \dfrac{m\Delta v}{\Delta t}

F = \dfrac{0.145(40-0)}{0.05}

F = 116 N

Force of the pitcher hand on the ball is equal to 116 N.

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Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the
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Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

\lambda=471\ nm

We need to calculate the distance between the two slits

m\times \lambda=d\sin\theta

d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

d = distance between the two slits

Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

d=40.11\times10^{-6}\ m

d = 40.11\ \mu m

Hence, The distance between the two slits is 40.11 μm.

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