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3 years ago

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If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp

eed of stanton's bat decrease due to the collision? (give your response as a positive value in units of miles per hour. assume that the bat and ball are moving in a single dimension. a baseball has a weight of 5 ounces and the bat stanton uses has a weight of 32 ounces.)

1 answer:

Mice21 [21]3 years ago

3 0

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

$m_1(v_f - v_i) = m_2(\Delta v)$

given that

$m_1 = 5 ounce$

$m_2 = 32 ounce$

For ball the change in speed will be

$v_f - v_i = (120 - 77)mph$

now from above equation

$5\times (120 - 77) = 32 \times \Delta v$

$\Delta v = 6.72 mph$

so speed of bat will decrease by 6.72 mph

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Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

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Where m is the mass of the pendulum.

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At the bottom of the swing,the height is zero, hence the potential energy is also zero.

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Energy can neither be created e destroyed. It can only be transferred from one form to another. Implying this law and the above explainations we conclude that at the bottom of the pendulum,the potential energy=0 and the kinetic energy=294J as the entire potential energy is converted to kinetic energy at the bottom.

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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$

Explanation:

From the question we are told that

The current on the copper is $I = 20 \ A$

The cross-sectional area is $A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2$

The number of copper atom in the wire is mathematically evaluated

$n = \frac{\rho * N_a}Z}$

Where $\rho$ is the density of copper with a value $\rho = 8.93 \ g/m^3$

$N_a$ is the Avogadro's number with a value $N_a = 6.02 *10^{23}\ atom/mol$

Z is the molar mass of copper with a value $Z = 63.55 \ g/mol$

So

$n = \frac{8.93 * 6.02 *10^{23}}{63.55}$

$n = 8.46 * 10^{28} \ atoms /m^3$

Given the 1 atom is equivalent to 1 free electron then the number of free electron is

$N = 8.46 * 10^{28} \ electrons$

The current through the wire is mathematically represented as

$I = N * e * v * A$

substituting values

$20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}$

=> $v = 0.0002808 \ m/s$

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