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3 years ago

Hi! I really need help on these questions, it would be great if you can answer them!

Physics

2 answers:

Nikitich [7]3 years ago

6 0

1: Equilibrium

[] When two forces are equal in opposite directions, they are in a state of equilibrium

2: Friction

[] It can be hard to sometimes slide a paper across a table, roll a car across the ground, etc etc. This is because of a force called friction which resists motion

3: Displacement

[] Work is only "done" when the object or surroundings are having movement, or in motion, and this creates displacement

4: Force

[] A force is a push/pull in any direction, this is a simple definition so not much can be added.

5: Inertia

[] Inertia is a force that likes to remain unchanged, or without movement

6: Effort

[] "Meg is holding up weight without movement. She hasn't done any work since it did not move, but is tired. Why?" Effort! She still put in effort even though no "work" happened

7: Work

[] Force times displacement is the equation for work

8: Motion

[] Look back to #3, when there is motion we can tell work is done

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Ilya [14]3 years ago

5 0

Equilibrium, friction, displacement, force, inertia

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hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .

fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)

do =original position = 2 m

vo = original VERTICAL velocity = 0

a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t - 1/2 ( 9.81)t^2

to show t = .639 seconds to hit the ground

During this .639 seconds it flies horizontally at 10 m/s for a distance of

10 m/s * .639 s = 6.39 m

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1 year ago

Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s

OleMash [197]

Answer:

My scenario would be A Car vs. a guard rail on a road. You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

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2 years ago

How high would you have to lift a 1000kg car to give it a potential energy of:

Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

It is mathematically expressed as:

P.E = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

h = $\frac{P.E }{mg}$

For 2.0 x 10³ J;

h = $\frac{2000}{1000 x 9.8}$ = 0.204m

For 2.0 x 10⁵ J;

h = $\frac{200000}{9.8 x 1000}$ = 20.4m

For 1.0kJ = 1 x 10³J;

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3 years ago

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Anvisha [2.4K]

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3 years ago

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Anit [1.1K]

Answer:

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 ) ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66² - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

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= 86.53 ft / s

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acceleration = - 28 ft /s²

v = u - at

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3 years ago