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Mrrafil [7]
3 years ago
5

A 250-gram cart starts from rest and rolls down an inclined plane from a height of 0.541m. Determine its speed at the bottom of

the incline.
Physics
1 answer:
Alex777 [14]3 years ago
8 0
<span>Δ PE = m g ( h1 - h2 ) = 0.25 * 9.8 ( 0.541 - 0.127 ) = 2.45 * 0.414 = 1.0143 J Δ PE = KE = 1 / 2 m V^2 1 / 2 m V^2 = 1.0143--- v^2 = 1.0143 * 2 / 0.25 = 8.1144 --- v= 2.8486 m/s</span>
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3 years ago
An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 f
____ [38]
<h2>a) Initial velocity = 83 ft/s</h2><h2>b) Object's maximum speed = 99.4 ft/s</h2><h2>c) Object's maximum displacement = 153.64 ft</h2><h2>d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e) The displacement is zero when t = 5.70 seconds</h2><h2>f) Object's maximum height = 153.64 ft</h2>

Explanation:

We have velocity

             v(t)= -32t + 83

Integrating

              s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

              46 = -16 x 0²+83 x 0+C

                 C = 46 feet

So displacement is

              s(t) = -16t²+83t+46

a) Initial velocity is

                 v(0)= -32 x 0 + 83 = 83 ft/s

       Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

             0 = -16t²+83t+46

             t = 5.70 seconds

Substituting in velocity equation

           v(t)= -32 x 5.70 + 83 = -99.4 ft/s

           Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

   That is

                 -32t + 83 = 0

                       t = 2.59 s

Substituting in displacement equation

                s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

   The displacement is zero when t = 5.70 seconds

f) Same as option d

   Object's maximum height = 153.64 ft

5 0
3 years ago
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