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earnstyle [38]
3 years ago
10

Describe the relationship between motion and a reference point.

Physics
2 answers:
Stells [14]3 years ago
8 0

Answer:

A <u>object </u>is in motion when its distance from another object it changing.

a <u>reference </u>point is a place or object used for comparison to determine if something is in motion.  You can assume that the refrence point is stuck or not moving.

steposvetlana [31]3 years ago
4 0

Answer:

An object is in motion when its distance from another object is changing. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.

Explanation:

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Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only
Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

v=u+at

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec²  [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2

\Rightarrow 16\times t^2=576

\Rightarrow t^2=\frac{576}{16}

\Rightarrow t^2=36

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

8 0
3 years ago
The figure shows a graph of electric potential versus position along the x-axis. A proton is originally at point A, moving along
Lostsunrise [7]
Can we see the figure?
5 0
3 years ago
Phineas and Ferb are riding their bikes on Main Street. They begin at 11 Mile Rd at
Lady bird [3.3K]

Answer:

1 hour and 25 minutes

Explanation:

hopefully this helps ,, but like WHY phineas and ferb hshjdshjdhdhdbd im so childish

7 0
3 years ago
PLEASE HELP ME<br> What is the net force acting on the car?
Hitman42 [59]
The net force acting on the car is 3 which slows the car down.
3 0
3 years ago
Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori
LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0
4 years ago
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