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sasho [114]
3 years ago
7

Displacement: Consider a deer that runs from point A to point B. The distance the deer runs can be greater than the magnitude of

its displacement, but the magnitude of the displacement can never be greater than the distance it runs. True or False?
Physics
1 answer:
adoni [48]3 years ago
4 0

Answer:

The given statement concerning distance and displacement is true.

Explanation:

If any point travels along any path between two points A and B the distance it travels is the length of the actual path it travels.

Mathematically the distance between two points is the algebraic sum of the infinitesimal distances that the particle travels.

Mathematically S=\int |ds|

Displacement by definition  is the shortest distance between the initial and final positions of the particle and is given by the straight line joining the two points.

Since a line is the shortest path that exists between 2 points thus it will be always smaller than any other path connecting the 2 points hence will be smaller than the distance.

While as if the object travels along a straight line without changing it's direction the distance in that case equals the displacement.

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A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

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2 years ago
Why is the sky blue?
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Temka [501]

Answer:3,600 Newtons

Explanation:

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3×10^3squared

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Substituting the above given values we get,F=(1500kg) (2.0m /s^2 squared)=3000 N=3×10^3 squared N.

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