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3 years ago

An object moves 2.5 m. This is an example of a _______.

Physics

2 answers:

makvit [3.9K]3 years ago

7 0

The correct answer is A. Distance

Hope this helps

Stells [14]3 years ago

6 0

An object moves 2.5 m. This is an example of a Distance

Answer: Option A

Explanation:

Direction denotes whether positive or negative or the left and right direction of an object. Speed and velocity are used to represent the distance travelled by the object with respect to the time taken by the object to travel.

Distance and displacement denotes the position the object has travelled from its previous location. Hence Option A is correct and B, C, D are irrelevant.

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A car traveling with 500,000 J of kinetic energy is brought to a kinetic energy of

allsm [11]

Answer:

33,333.33 N

Explanation:

Given that :

Initial kinetic energy = 500,000 J

Final kinetic energy = 100,000 J

Using the relation :

Force * time = change in momentum (Newton's law)

Force (F) * 0.12 = (500,000 - 100,000)

0.12F = 400,000 J

Force = (400,000 J) / 0.12s

Force = 33333.333

Force = 33,333.33 N

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3 years ago

While performing an experiment involving a reaction between two chemicals, the scientist observes that the reaction container ha

AVprozaik [17]

When an object is becomes warm its an example of exothermic reaction.

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3 years ago

There is a block on your desk that

Anni [7]

Explanation:

volume=3x4x6=72cm³

density=mass/volume

so:

184.32/72=2.56g/cm³ for the block

7 0

3 years ago

Read 2 more answers

Find equailent resistance

vovikov84 [41]

Answer:

0.71ohms

Explanation:

AC and CB are in series hence the total resistance is 3+2=5ohms

AD and DB are in series hence the total resistance is 3+2=5ohms

These resistances are parallel to resistance AB

Let the equivalent resistance be R.

For resistances in parallel

Hence the total resistances are:

5//1//5

1/R =1/5 + 1/1 +1/5 = 1+5+1 /5 = 7/5

Hence the total resistance is R= 1/ 7/5 = 1×5/7= 5/7=0.71ohms

4 0

3 years ago

At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc

amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

$F=At^2$

Here F = 888.93 N when t = 1.25 s

$F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2$

The value of A is $568.91\ N/s^2$ .

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

$J=\int\limits {F{\cdot} dt}$

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

$J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns$

(b) Impulse is also equal to the change in momentum as :

$J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s$

Hence, this is the required solution.

5 0

3 years ago