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Marta_Voda [28]
3 years ago
15

for a moving object distance covered by it is always greater than or equal to the displacement of the object in a given time. ex

plain.​
Physics
1 answer:
AleksandrR [38]3 years ago
6 0
<h3><u>Answer</u></h3>

  • Distance is equal to the Total Distance covered by a body, from the initial till the final point.

  • Displacement is equal to the shortest distance between two points.

  • So we known that Distance can only be equal to or greater than the displacement and can never be shorter than the displacement.

  • This is just common sense how can anything be shorter than the shortest path itself. But it can be equal to the shortest path
<h3>━━━━━━━━━━━━━━</h3>

<h3><u>Know </u><u>More</u></h3>

☯ Distance is a scalar quantity and has only magnitude but no direction.

☯ Displacement is a vector quantity and has both magnitude and direction.

☯ Distance can only have +ve values whereas displacement can be +ve, -ve or even be zero.

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URGENT!!!!!!!<br><br> PLEASE HELP WITH THIS PHYSICS PROBLEM
Levart [38]

Explanation:

Let

x_1 = distance traveled while accelerating

x_2 = distance traveled while decelerating

The distance traveled while accelerating is given by

x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2

\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2

\:\:\:\:\:= 1125\:\text{m}

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

v^2 = v_0^2 + 2ax_2

0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2

Solving for x_2, we get

x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}

\:\:\:\:\:= 4327\:\text{m}

Therefore, the total distance x is

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3 0
3 years ago
The amount of WORK done is determined by 2 factors.
RUDIKE [14]

Explanation :

Work is done when a force is applied to create a displacement on an object.

Thus, the work done depends on the two factors i.e.

(1) Applied force (F)

(2) Distance or displacement (d)

Mathematically, work done is W= F.s

It also depends on the angle between the force and the displacement.

        W=Fs\ cos\ \theta

For example,

A person carries a weight of 20 kg and lifts it on his head 1.5 m above the surface. So, the work done by him on the luggage will be:

W= F\times s

or

W=m\times g \times s

W=20\ kg\times 9.8\ m/s^2\times 1.5\ m

So, W=294\ Joules

Hence, the work done by him on the luggage is 294 Joules.

6 0
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