Yes, an object<span> that was set in motion in the past by some force, but that is no longer being acted on by a net force, is </span>moving<span> but with </span>zero acceleration<span>, i.e. it is </span>moving<span> at constant velocity.</span>
I think its 13...........
<span>3.78 m
Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes.
7.2 m/s / 9.81 m/s^2 = 0.77945 s
The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving
d = 1/2 A T^2
d = 1/2 9.81 m/s^2 (0.77945 s)^2
d = 4.905 m/s^2 0.607542 s^2
d = 2.979995 m
So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height.
d = 2.979995 m + 0.8 m = 3.779995 m
Rounding to 2 decimal places gives us 3.78 m</span>
Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.
There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:
(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2
We can reduce this to:
(15 x 10)m/36 s^2 = 150 m/36 s^2
Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
efficiency = (useful energy transferred ÷ energy supplied) × 100
It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:
<u>Way #1:</u>
We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.
(useful energy) / (energy supplied) = (175J) / (250J) = <em>70% efficiency</em>
================================
<u>Way #2: </u>
How much of the energy is wasted ? . . . 75J wasted
What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted
30% of the input energy is wasted. That leaves the other <em>70%</em> to be useful energy.
