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dangina [55]
3 years ago
14

Why must one use a reference point to determine whether or not an object is in motion

Physics
1 answer:
GaryK [48]3 years ago
7 0
Because there's no such thing as "really" moving. 
ALL motion is always relative to something.

Here's an example:
You're sitting in a comfy cushy seat, reading a book and listening
to your .mp3 player, and you're getting drowsy.  It's so warm and
comfortable, your eyes are getting so heavy, finally the book slips
out of your hand, falls into your lap, and you are fast asleep.

-- Relative to you, the book is not moving at all.
-- Relative to the seat, you are not moving at all.
-- Relative to the wall and the window, the seat is not moving at all.
-- But your seat is in a passenger airliner.  Relative to people on the
    ground, you are moving past them at almost 500 miles per hour !
-- Relative to the center of the Earth, the people on the ground are moving
   in a circle at more than 700 miles per hour.  
-- Relative to the center of the Sun, the Earth and everything on it are moving
   in a circle at about 66,700 miles per hour ! 

How fast are they REALLY moving ?
There's no such thing.
It all depends on what reference you're using.

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What type on msd causes inflammation of the tendons that correct bones to muscles
svp [43]

I believe it would be Tendonitis

8 0
3 years ago
What does a force moving an object through a distance transfer? A. sound B. energy C. weight D. mass
Ulleksa [173]

Answer:

it transfers energy as it moves

4 0
3 years ago
If a car that is moving 20.0 m/s has a momentum of 29000 kg·m/s, what mass is the car?
Strike441 [17]

Answer:

Mass = 1450kg

Explanation:

P = M * V (where p is momentum, m is mass and v is velocity)

29000 = 20 * M

M = 29000 / 20

M= 1450 kg

3 0
3 years ago
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
4 years ago
In metals, some electrons from the outer shells of atoms are shared with other atoms, with the result that electrons can freely
Kazeer [188]
<h2>Answer:Option F</h2>

Explanation:

Metals have few number of electrons in their valence shell.

The force with which the nucleus of the metal atom is pulling the electron towards itself is lower on the valence shell because the distance between the nucleus an valence shell is large.

So,there are a lot of free electrons in metals.

When the ends of metals are at different potential,electrons flow from one end to another to balance the potential.This feature of metals allows them to conduct electricity.

Similarly,when the ends of metals are at different temperatures,conduction will take place and electrons are the carriers of conduction.This feature of metals allows them to conduct heat.

So,metals are good conductors of heat and electricity.

4 0
4 years ago
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