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Andru [333]
3 years ago
15

HELP WITH THESE! IVE ASKED 5 TIMES!

Physics
1 answer:
Sedbober [7]3 years ago
7 0

3). Acceleration = (change in speed) / (time for the change)

Why, that's just the slope of the speed graph. Looks like about 0.6 m/s² to me.

5). As I heard someone say just a moment ago: "Acceleration is just the slope of the speed graph". The slope of the graph in #5 is negative, and it's constant. So the acceleration is negative and constant.

4). The instantaneous accelerating at any point on this graph is the slope of the graph AT THAT POINT. It changes at different places on this graph, BUT, by the time it gets to 10 sec, the graph is pretty much a straight line, so you can get pretty close to estimating its slope right there in that small region. From 10-11 sec, I see the speed rise from 14 to 16 m/s, so the acceleration in that little area is 2 m/s² .

7). The speed is decreasing as time goes on, so the acceleration is negative. The graph has different slope in different places, so the acceleration is changing.

9). The speed is increasing as time goes on, so the acceleration is positive. The graph has different slope in different places, so the acceleration is changing.

last problem, no number). They want you to find a place on the speed graph where the acceleration is negative. That means a place where the speed is decreasing, so it has to be somewhere before 4 seconds ... after 4 seconds, the speed starts to increase as time goes on.

So now you just have to find a place on the graph, before 4 seconds, where the speed drops by 2 in the time of one second. I couldn't blow it up enough to work with it. It looks to me like it's about 1.5 seconds, but you'll have to check it out.

The reason I'm guessing 1.5 seconds:

-- at 1.5 sec, the speed is 3 m/s

-- 1/2 sec earlier, at 1.0 sec, the speed is 4 m/s

-- 1/2 sec later, at 2.0 sec, the speed is 2 m/s

So around 1.5 sec, you can see a place on the graph that's almost straight, and the speed decreases by 2 m/s in the space of 1 second. So the average acceleration from 1s to 2s is 2 m/s², and 1.5s is right in the middle of that period of time.

Oh phoo ! I didn't even look at the choices until just now and 1.5s isn't one of them. You ought to be able to handle them now. Remember, it has to be before 4 seconds, because the acceleration is all positive after that.

(I know this is more detailed than your class is expecting, and that's why my answer isn't one of the choices. There's a simple reason for my getting too complicated. You see, I've sweated through so many years of Calculus and Differential Equations that my brain got fried, and now, when I work on middle-school and high-school problems, it HAS TO go through those channels in my brain. There's no other way.)

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A parallel-plate capacitor is constructed of two horizontal 15.6-cm-diameter circular plates. A 1.4 g plastic bead with a charge
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Answer: The upper plate is positive charged. b) 0.53 µC.

Explanation:

As the plastic bead is suspended between the two plates, the external forces acting upon it must be equal and opposite each other.

The external forces acting upon it, are gravity (always downward) and the on due to electric Field, which must aim upwards, to keep the system in equilibrium.

As the charge on which the force must be act is a negative charge, this means that the force and the electric field must aim in opposite direction, so the electric field must point from the upper plate to the lower one.

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b) In order to know the charge on the positive charged plate, we know that the capacitance , by definition, is as follows:

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It can be showed, that for a parallel plate capacitor, the capacitance can be expressed as a relationship of geometric parameters, as follows:

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Q = ε₀ A E = 8.85. 10⁻¹² farad/m. π (0.156)2 . ¼. 3.12. 10⁶ V/m = 0.53 µC  

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