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Delicious77 [7]
3 years ago
5

Marisa’s car accelerates at an average rate of 2.6m/s^2. Calculate how long it takes her car to accelerate from 24.6m/s to 26.8m

/s? Show your work.
Physics
1 answer:
djverab [1.8K]3 years ago
8 0

given info is... Acceleration(a)=2.6m/s^2

                       final velocity(v)=26.8m/s

                       initial velocity(u)=24.6m/s

need to find.... time(t)=?

a=\frac{v-u}{t} \\2.6=\frac{26.8-24.6}{t} \\\\

t=\frac{v-u}{a}

t=\frac{26.8-24.6}{2.6}

t=0.846s

Explanation:

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A 0.380 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 28.0 N/
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Initial momentum of the Play-doh: 0.0600 x 2.60 = 0.156 kg/m/s

Total mass of the block and play-doh: 0.38 + 0.0600 = 0.44 kg.

Final momentum is mass x velocity = 0.44v

V = Initial momentum / mass

V = 0.156 / 0.44 = 0.3545 m/s

Work done by spring is equal to the Kinetic enrgy.

Work Done by spring = 1/2 *28.0 * distance^2 = 14 * d^2

KE = 1/2 * 0.44* 0.3545^2

set to equal each other:

14 * d^2 = 0.22 *0.12567

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You throw a 20 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 25.0 m/
siniylev [52]

Answer:

a) 30.3 m/s b) 46.8 m

Explanation:

a)

  • Assuming no friction present, the work-energy theorem says that the work done on the rock by the only force acting on it (gravity force, aiming downward), is equal to the change of kinetic energy of the rock, at any point.
  • When it's at 15.0 m high, the work done by gravity on the rock can be written as follows:

        W = m*g* h* cos 180\º = -20 N* 15.0 m = -300 J (1)

  • The change in kinetic energy of the rock can be expressed in this way:

        \Delta K = K_{f} - K_{0}  = \frac{1}{2} * m* (v_{f}^{2} -v_{0} ^{2} ) (2)

  • where m= 2.04 kg, vf = 25.0 m/s.
  • As (1) and (2) are equal each other, we can solve for v₀, as follows:

        = -300 J = 0.5*2.04 kg *(( 25.0 m/s)^{2} -v_{0}^{2} )\\ \\ v_{0}^{2} = \frac{300*2 J}{2.04kg} + 625 (m/s)2 = 894. 1 (m/s)2 \\ \\  v_{0} = \sqrt{894.1 (m/s)2} = 30.3 m/s

b)

  • When the rock arrives to its highest point, it will momentarily come to a stop, before changing direction to start falling.
  • So, the final kinetic energy, will be zero.
  • Applying the work --energy theorem again, we can write the following equation:

        \Delta K = K_{f} - K_{0}  =( 0- \frac{1}{2} *m*v_{0}^{2}) = -m*g*h_{max}

  • Replacing by v₀ = 30.3 m/s, rearranging and simplifying common terms, we can solve for hmax, as follows:

        h_{max} = \frac{v_{0} ^{2} }{2*g} =\frac{(30.3 m/s)^{2}}{2*9.8 m/s2} = 46.8 m

7 0
3 years ago
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