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3 years ago

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Anissa slides down a playground slide sloped at 25o. The coefficient of kinetic friction between Anissa and the slide is 0.15. I

f Anissa’s mass is 55 kg, how fast is she going 2.0 s after starting from rest?

1 answer:

Keith_Richards [23]3 years ago

4 0

Answer:5.62 m/s

Explanation:

Given

Inclination $\theta =25^{\circ}$

Coefficient of kinetic Friction $\mu _k=0.15$

mass of Anissa's $m=55 kg$

time interval $t=2 s$

acceleration of the system

$a=g\sin \theta -\mu _kg\cos \theta$

$a=9.8\sin (25)-0.15\cdot 9.8\cos (25)$

$a=4.14-1.33=2.81 m/s^2$

$v=u+at$

where v=Final Velocity

u=Initial Velocity

a=acceleration of system

t=time

$v=0+2.81\times 2$

$v=5.62 m/s$

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

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So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

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A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of th

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Answer:

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Explanation:

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As we know that the frequency is inversely proportional to the wavelength of the velocity is same.

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The frequency of the second wave is half of the frequency of first one.

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Answer:

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