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cluponka [151]
2 years ago
11

A puck of mass m moving at speed v on a horizontal, frictionless surface is stopped in a distance d because a hockey stick exert

s an opposing force of magnitude F on it. If the stopping distance d increases by 58 %, by what percent does the average force needed to stop the puck change, assuming that m and v are unchanged?
Physics
1 answer:
MArishka [77]2 years ago
8 0

Answer:

-36.71%

Explanation:

Using the Generalized Work Energy Principle, the puck is brought to rest by an external force and the system has no kinetic energy

U_i+W=U_f\\K_i+W=K_f\\\\0.5mv_i^2+F\bigtriangleup xcos\theta=0J\\\\0.5mv_i^2+F\bigtriangleup xcos180\textdegree=0J\\\\0.5mv_i^2=F\bigtriangleup x\\\\F=\frac{0.5mv_i^2}{2\bigtriangleup x}

#Denote the stopping distance and force required with a prime and observe from (i) that:

F\infty\frac{1}{\bigtriangleup x}\\\\F\prime =F.\frac{\bigtriangleup x}{(\bigtriangleup x)\prime}\\\\  #If stopping distance increases by a factor of 79/50;

F\prime=F.(79/50)^-^1\\\\F\prime=\frac{50F}{79}\\\\\therefore \bigtriangleup F=F\prime -F\\\\\bigtriangleup F=\frac{50F}{79}-F=-\frac{29F}{79}

So F decreases by 36.71% . we expect force to reduce since the same amount of work is repeatedly done on the system over a long distance.

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ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

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2 years ago
Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter
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The frequency of the radio wave is:
f=102.1 MHz = 102.1 \cdot 10^6 Hz

The wavelength of an electromagnetic wave is related to its frequency by the relationship
\lambda= \frac{c}{f}
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Barker is unloading 20kg bottles of water from this delivery truck when one of the bottles tips over and slides down the truck r
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2 years ago
A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al
anzhelika [568]

Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

          vₓ = v₀ₓ = 1.10 m / s

          aₓ = 0

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initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

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          ay = (v_{oy} -v_{y}) / t

          ay = (0 -10.9) / 1.6

          ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

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