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GaryK [48]
3 years ago
11

Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,

Zak decides to get a running start and then slide across the floor. Part A If Zak's speed is 3.00 m/s when he starts to slide, what distance d will he slide before stopping? Use 9.81 m/s2 for the acceleration due to gravity.
Physics
1 answer:
insens350 [35]3 years ago
3 0

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction \mu_{k} = 0.250

Zak's speed v = 3 \frac{m}{s}

Gravitational acceleration g = 9.8 \frac{m}{s^{2} }

Work done by frictional force is given by,

  W = \Delta K

 \mu _{k} mg d = \frac{1}{2} m v^{2}

  d = \frac{v^{2}  }{2 g \mu _{k} }

  d = \frac{9}{2 \times 9.8 \times 0.250}

  d = 1.84 m

Therefore, the distance travel before stopping is 1.84 m

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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(
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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})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

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:

v=-256ft/s

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3 years ago
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The following are connected in parallel circuit at home EXCEPT:
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2 years ago
A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts
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Answer:

4987N

Explanation:

Step 1:

Data obtained from the question include:

Mass (m) = 0.140 kg

Initial velocity (U) = 28.9 m/s

Time (t) = 1.85 ms = 1.85x10^-3s

Final velocity (V) = 37.0 m/s

Force (F) =?

Step 2:

Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:

F = m(V + U) /t

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F = 4987N

Therefore, the magnitude of the horizontal force applied is 4987N

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3 years ago
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