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3 years ago

5.) Scott pushes a piano along a slope and has an initial velocity of 10.0 m/s [up]. Its acceleration is 2.0 m/s2 [down].

Physics

1 answer:

forsale [732]3 years ago

5 0

Answer:

a) 6m/s, 0m/s, -6m/s

b) 16m, 25m, 16m

Explanation:

For the first part of the problem, use the Kinematic Equation for Final Velocity:

You are looking for the final velocities of the times: 2s, 5s, and 8s (plug those in for t). We also know that the acceleration is going down so it is negative (-2 is plugged in for a). 10 is the initial velocity and it is positive.

For the second part of the problem, use the Kinematic Equation for Displacement:

You are looking for the displacement of the piano at times: 2s, 5s, and 8s (plug in and square at t). Again, the acceleration is negative because it is in the direction of down, so plug -2 in for a. The initial velocity is positive 10. Plugging those numbers in for each variable should get the correct answer.

Hopefully this helped, I went on here in search of an answer and found it out myself and since no one has answered it yet, I'll help someone else out.

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Answer:

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3 years ago

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

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The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

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4 years ago

A cyclist racing in the keiran is riding at the top of the track at 5m/s. Then he sprints downhill in the sprinting to the finis

kumpel [21]

The cyclist is moving by uniformly accelerated motion, with an initial velocity of $v_i=5~m/s$ and an acceleration of $a=9~m/s^2$ .
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$a= \frac{v_f-v_i}{\Delta t}$
where $v_f$ is the final velocity and $\Delta t$ is the time between the end and the beginning of the motion, and in our case it is 1.75 s. Therefore, from this relationship we can find the final velocity:
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