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3 years ago

If anyone could answer my question , please help -,- xD

2 answers:

6 0

Kinematics is the branch of mechanics concerned with the motion of objects without the reference to the forces which cause the motion.

Paraphin [41]3 years ago

5 0

Kinematics is the study of motion, without any reference to the forces that cause the motion. It basically means studying how things are moving, not why they're moving. It includes concepts such as distance or displacement, speed or velocity, and acceleration, and it looks at how those values vary over time.

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Will mark as brainliest if correct!!!!!!!!!!!!!!!!!!!!!!!!!

Digiron [165]

Answer:

Incident ray

Explanation:

"To describe the reflection of light, we will use the following terminology. The incoming light ray is called the incident ray. The light ray moving away from the surface is the reflected ray. The most important characteristic of these rays is their angles in relation to the reflecting surface."

https://www.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-03

3 0

4 years ago

Read 2 more answers

a ball is dropped and falls with an acceleration of 9.8m/s^2 downward. it hits the ground with a velocity of 49m/s downward. how

yaroslaw [1]

The answer below...........

8 0

3 years ago

A compound whose molecules contain one boron atom and three fluorine atoms would be named monoboron fluoride. Please select the

Slav-nsk [51]

I know it is false...............

5 0

3 years ago

The Earth's Radius is 6.3710x106 m and mass is 5.9742x1024 kg. What is the acceleration due to gravity at Mount Everest (elevati

Shalnov [3]

Answer is

9.773m/s^2

-----------------------------------------------------------------------------

Given,

h=8848m

The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.

g′=g(1 − 2h/h)

=9.8(1 - 6400000/17696)

=9.8(1 − 0.00276)

9.8×0.99724

=9.773m/s^2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2

-----------------------------------------------------------------------

hope this helps :)

3 0

3 years ago

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

5 0

3 years ago