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2 years ago

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

1 answer:

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.

You might be interested in

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

2 years ago

A person wearing a shoulder harness can survive a car crash if the acceleration is smaller than -300 m/s . assuming constant acc

mars1129 [50]

To solve this problem, we use the equation:

where,

d = distance of collapse

v0 = initial velocity = 101 km / h = 28.06 m / s

v = final velocity = 0

a = acceleration = - 300 m / s^2

d = (-28.06 m / s)^2 / (2 * - 300 m / s^2)

3 0

2 years ago

B)A man walks 95 km, East, then 55 km, north. Calculate his RESULTANT

Varvara68 [4.7K]

The resultant displacement of the man is 109.77 km in the direction N60°E.

<h3>Displacement</h3>

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

<h3>Resultant displacement of the man </h3>

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

Then, y = 90 - 30

y = 60°

Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

Learn more about displacement at: brainly.com/question/321442

8 0

2 years ago

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

Rzqust [24]

Answer:

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

= 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0

2 years ago

If 2 grams of element X combine with 4 grams of element Y to form compound XY, how many grams of element Y would combine with 46

Butoxors [25]

92 grams of Y would combine with 46 grams of X

3 0

2 years ago