All categories

3 years ago

Measure if how far an object has moved.

1 answer:

5 0

You can't really measure how far an object has moved. If you weren't
watching it the whole time, you can only measure how far it IS now from
where it started, but you don't know what route it traveled to get there.

The distance between where it started and where it ended up is called
the object's "displacement". That's the length of the straight line between
those two points. And it's also the shortest possible distance the object
could have moved in order to get to where it is now.

Funny thing: When you walk all the way around a yard, a track, or a building,
or drive a car one lap around the track, your displacement is zero, because
you end up in the same place you started from, and the distance is zero.
If somebody saw you before and after, but didn't see you walk or drive,
they wouldn't know that you had moved at all.

You might be interested in

Tell me about your favorite science experiment. Give me a detailed account of the experiment and what you like about it.

Aleks [24]

Answer:

nngh have

bjruh hjrhhj be rnrnnrnrnnnrnjrjnnnnnnnnrnrn n n and I was nrn

Explanation:

jbbbbbhhhjjnnnnnnnnnnjhvcc

Explanation:

5 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago

Mickey had three substances. He hit each of them with a hammer. A red

Verizon [17]

Answer:

A & D

Explanation:

6 0

3 years ago

1. The illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the source?

Crank

Answer

1) A. 96 Candelas

2) A. Both of these types of lenses have the ability to produce upright images.

3) C. 5 meters

Explanation

The formula for calculation the luminous intensity is;

Luminous intensity = illuminance × square radius

Lv = Ev × r²

= 6 × 4²

= 6 × 16

= 96 Candelabra

For converging lenses, an upright image is formed when the object is between the lens and the principal focus while a diverging lens always forms and upright image.

A. Both of these types of lenses have the ability to produce upright images.

Luminous intensity = illuminance × square radius

square radius = Luminous intensity/ illuminance

r² = 100/4

= 25

r = √25

= 5 m

5 0

4 years ago

Read 2 more answers

Veolocity increases with

Aneli [31]

Velocity increases with speed

8 0

3 years ago