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9966 [12]
3 years ago
8

slader A baseball player slides to a stop on level ground. Using energy consideration, calculate the distance the 65 kg baseball

player slides, given his initial speed is 6 m/s and the force of friction against him is a constant 450N.
Physics
1 answer:
stich3 [128]3 years ago
6 0

Answer:

2.6 m

Explanation:

From work energy theorem we deduce that

0.5mv^{2}=fd

The kinetic energy equals to the work dine against friction energy.

Where m represent mass, v is the velocity, f is the frictional force and d is unknown distance that the player slides.

Making d the subject of the formula then d=\frac {mv^{2}}{2f}

Then by substituting 450N for f, 65 kg for m and 6 m/s for v we obtain

d=\frac {65*6^{2}}{2*450}=2.6m

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Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

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where :

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the car, with

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Soolving:

W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J

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3 years ago
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen
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Answer:

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c) Δy = 0.64 m

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a) and b) We apply the equation to the first laser

          λ = d / 20

          d y / x = m d / 20

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The second laser

        λ = d / 15

          d y / x = m d / 15

          y = m x / 15

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