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3 years ago

Pls guys what is displacement

Physics

2 answers:

insens350 [35]3 years ago

7 0

Explanation:

Displacement is a difference 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 displaced, or changed, by exactly the volume of that object.

Example: 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 displacement 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 displacement 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 student performs an experiment in measuring the period of a simple pendulum of known length 49.0 cm.He performed five trials a

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Correct Answer is Bb

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3 years ago

How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad

Rzqust [24]

initial angular speed is given by 33.3 rpm

$w_0 = 2\pi \frac{33.3}{60}$

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final angular speed is given by 78 rpm

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$w_f = 8.17 rad/s$

now by using kinematics we will have

$w_f = w_0 + \alpha * t$

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3 years ago

A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your

Nataly [62]

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

$1.66 = 0 + 0t + \frac{1}{2}9.8t^2$

$1.66 = 4.9t^2$

$\frac{1.66}{4.9} = t^2$

$\sqrt{0.339} = t\\ t = 0.582s$

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

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3 years ago

What determines the atomic number of an atom?

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The atom's atomic number is the number of protons in its nucleus.

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4 years ago

What is the potential difference between a point 0.48 mm from a charge of 2.9 nc and a point at infinity?

Nuetrik [128]

The potential at a distance r from a charge Q is given by
$V(r) = k_e \frac{Q}{r}$
where ke is the Coulomb's constant.

The charge in our problem is $Q=2.9 nC=2.9 \cdot 10^{-9} C$ ; for the point at $r=0.48 mm=0.48 \cdot 10^{-3} m$ , the potential is
$V_1 = k_e \frac{Q}{r}= (8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.9 \cdot 10^{-9} C}{0.48 \cdot 10^{-3} m}= 5.43 \cdot 10^4 V$

For the point at infinity, we immediately see that the potential is zero, because $r= \infty$ and so $V_2 = 0$ .

Therefore, the potential difference between the two points is
$\Delta V = V_1 - V_2 = V_1 = 5.43 \cdot 10^4 V$

6 0

3 years ago

Pls guys what is displacement​

Pls guys what is displacement