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liubo4ka [24]
3 years ago
9

Pls guys what is displacement​

Physics
2 answers:
insens350 [35]3 years ago
7 0

Explanation:

Displacement is a <em>difference </em>between an initial value and a final value, not factoring in any changes that happen in between. For example you let an object of a certain volume sink to the bottom of a container of water, the water level will be <em>displaced</em>, or changed, by exactly the volume of that object.

<em>Example</em>: A marble with volume of 3 cm³ is dropped into a graduated cyllinder with 10 mL of water. The water level rises to the 13 mL mark, so the <em>displacement </em>of the water by the marble is exactly 3 mL.

This doesn't just apply to a change in volume: If I move a box 3 meters along an axis in the positive direction, we say its <em>displacement</em> along that axis is 3 meters. If I moved it in the negative direction the same distance, the displacement would be -3 meters.

KengaRu [80]3 years ago
4 0

Answer:

Displacement is the shortest distance from your initial position to your final position. Therefore, displacement is a measurement of distance and not an "end-point."

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A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug
Vaselesa [24]

Answer:

1.57 * 10^{3} Q

Explanation:

The volume charge density is defined by ρ = \frac{Q}{V} (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are \frac{Coulombs}{m^{3} }, so we have to express the radius in meters:

inner radius = 4 cm * \frac{1 m}{100 cm} = 0.04m

outer radius = 6 cm * \frac{1m}{100cm}  = 0.06m

Now, we know that the volume of the sphere is calculated by the formula:

V = \frac{4}{3}\pi r^{3}, and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = \frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3}), where r_{o} is the outer radius and r_{i} is the inner radius.

Replacing the volume formula in the Equation A:

ρ = \frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}

ρ = \frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }

Replacing the values of the outer and inner radius whe have:

ρ = \frac{3Q}{4\pi (1.52 * 10^{-4})}

ρ = 1.57 * 10^{3} Q

4 0
4 years ago
Name two elements that have similar properties. How can you tell using the periodic table?
tangare [24]
Helium and nitrogen because they are both from the noble gas group.
7 0
4 years ago
9. What could cause scientist to change their mind?
MakcuM [25]
A if it is one answer and if it’s multiple choice A and D
6 0
2 years ago
HELP WITH THESE! IVE ASKED 5 TIMES!
Sedbober [7]

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.)

7 0
3 years ago
A crane lifts an l-bean to a height of 30 meters giving it a potential energy of 108,486). What is the mass of the I-beam?
andrew-mc [135]

369 kg

Explanation:

Step 1:

We know that the expression for potential energy is

PE = mgh

where, m =Mass

            g = Acceleration due to gravity

            h Height

Step 2:

Here m = ?

         PE = 108486

         h = 30 m

          g = 9.8 ms⁻²

108486 = 30 * 9.86 * m

m= 108486 / (9.8*30) = 369 kg

5 0
3 years ago
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