Physics, 09.10.2019 22:30, nisslove

The fastest pitched baseball was clocked at 47 m/s. assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g. a) draw a free-body diagram of the ball during the pitch. the ball is moving to the right.
b) what force did the pitcher exert on the ball during this record-setting pitch?
c) estimate the force in part b as a fraction of the pitcher’s weight. assume that the pitcher has a mass of 84 kg.

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answered: lizzbugg9880

The force the pitcher exert on the ball during this record-setting pitch is 160.15 N.

The force exerted on the baseball as a fraction of the pitcher's weight is 0.195.

The given parameters;

initial velocity of the baseball, u = 47 m/sdistance traveled by the ball, d = 1 mmass of the baseball, m = 145 g = 0.145 kg

The free-body diagram of the ball during the pitch is in the image added.

The acceleration of the baseball over the given distance is calculated as;

v² = u² - 2as

when the ball stops, the final velocity, v = 0

0 = u² - 2as

2as = u²

$a = \frac{u^2}{2s} \\\\a = \frac{(47)^2}{2(1)} \\\\a = 1,104.5 \ m/s^2$

The force the pitcher exert on the ball during this record-setting pitch is calculated as;

F = ma

F = 0.145 x 1,104.5

F = 160.15 N.

The weight of the pitcher is calculated as;

W = mg

W = 84 x 9.8

W = 823.2 N

The force exerted on the baseball as a fraction of the pitcher's weight is calculated as follows;

$n= \frac{160.15}{823.2} = 0.195$

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The fastest pitched baseball was clocked at 47 m/s. assume that the pitcher exerted his force (assum

answered: persondontknowhelp

A) In attachment

Find the free-body diagram in attachment. There are only two forces acting on the ball during the pitch:

- The force applied by the pitcher, F, which acts in the horizontal direction

- The force of gravity acting on the ball, also called weight, whose magnitude is

W=mg

where m is the mass and g is the acceleration of gravity, and it acts in the downward direction.

B) 160.2 N

First of all, we need to find the acceleration of the ball. We can do it by using the SUVAT equation

v^2-u^2=2ad

where

v = 47 m/s is the final velocity of the ball

u = 0 is the initial velocity (the ball starts from rest)

a is the acceleration

d = 1.0 m is the distance covered by the ball while the force is applied

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 m/s^2$

And since we know the mass of the ball,

m = 145 g = 0.145 kg

We can find the force exerted on it by using Newton's second law:

F=ma=(0.145)(1104.5)=160.2 N

C) 0.195 times the pitcher's weight

The weight of the pitcher is given by

W = mg

where

m = 84 kg is the mass of the pitcher

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

Solving,

W=(84)(9.8)=823.2 N

So now we can express the force found in part B) as a fraction of the pitcher's weight:

$\frac{F}{W}=\frac{160.2}{823.3}=0.195$

answered: Guest

force required to detach it is given by

$f = 4 * 10^{-8} n$

charges are taken to be fundamental charge

$q_1 = q_2 = 1.6 * 10^{-19}c$

now we have

$f = \frac{kq_1q_2}{r^2}$

$4 * 10^{-8} = \frac{9*10^9 * 1.6*10^{-19}* 1.6* 10^{-19}}{r^2}$

$r = 7.6 * 10^{-11} m$

so above is the minimum distance at which it will detach

answered: Guest

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The fastest pitched baseball was clocked at 47 m/s. assume that the pitcher exerted his force (assum...