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Butoxors [25]
3 years ago
15

The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the t

ime of impact is​
Physics
1 answer:
Amanda [17]3 years ago
7 0

Answer:

The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is​ extremely small.

Explanation:

During the collision between two particles, a large force acts between the two colliding bodies for a relatively short time, also known as impulse.

Therefore, the principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is​ extremely small.

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In a given system of units the ratio of the unit of volume to that of area gives the unit of​
ch4aika [34]

Answer:

length

Explanation:

SI unit of volume = m^3

SI unit of area = m^2

volume unit / Area unit = m^3 / m^2

i.e, unit of length

8 0
3 years ago
The force between charged objects decreases when their separation ..... A)increase<br> B)decrease
weqwewe [10]
Answer A: When their separation increases.

Hope this helps! :D
6 0
3 years ago
You make the following measurements of an object 42kg and 22m3 what would the objects density be
miv72 [106K]
The density of the object is approximately 1.91 kg per m³.

42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship $$density = mass / volume$$ to solve for the object's density.

42 kg \div 22 m³ \approx 1.91 kg per m³.
4 0
3 years ago
Read 2 more answers
Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict
GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = m

Mass of the heavier cart = 2m

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ KE =  \frac{mv^2}{2}

⇒ KE =  \frac{mv^2}{2}\times \frac{m}{m}

⇒ KE =  \frac{m^2v^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(mv)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ K=\frac{(F\times t)^2}{2m}           ...equation (i)

KE (K1) of mass 2m.

⇒ K_1=\frac{(F\times t)^2}{2\times 2m}

⇒ K_1=\frac{(F\times t)^2}{4m}         ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ \frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}

⇒ \frac{K_1}{K} =\frac{2m}{4m}

⇒ \frac{K_1}{K} =\frac{2}{4}

⇒ \frac{K_1}{K} =\frac{1}{2}

⇒ K_1=\frac{K}{2}

⇒ K_1=\frac{1}{2}K

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0
3 years ago
What is the area
bearhunter [10]

Answer:e3

explanation: the answer is 200

4 0
3 years ago
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