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3 years ago

I WILL MARK BRAINLIST!!!

Physics

2 answers:

Anna35 [415]3 years ago

7 0

Answer:

Mary is correct because as the liquid cools, the molecules slow down as they lose kinetic energy.

Explanation:

The way to identify the temperature of a body, according to common sense, is to use the words hot and cold, but in fact, hot is when the molecules are very agitated, causing an increase in body temperature. The cold is related to the more stationary state of the molecules, the less agitated, the colder the body becomes.

With that, we can say that as the liquid cools, the molecules move less and less, which decreases their kinetic energy, which is the energy related to movement. In this case, Mary is correct because, as the liquid cools, the molecules slow down as they lose kinetic energy.

svetoff [14.1K]3 years ago

5 0

Answer:

Mary is correct because as the liquid cools, the molecules slow down as they lose kinetic energy.

Explanation:

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The following graph shows the force exerted on and the displacement of object being pulled

Tomtit [17]

The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.

<h3>Work done by the applied force</h3>

The area under force versus displacement graph is work done.

The total work done by pulling the object 7 m, can be grouped into two areas;

First area, A1 = area of triangle from 0 m to 2.0 m
Second area, A2 = area of trapezium, from 2.0 m to 7.0 m

A1 = ¹/₂ bh

A1 = ¹/₂ x (2) x (20)

A1 = 20 J

A2 = ¹/₂(large base + small base) x height

A2 = ¹/₂[(7 - 2) + (7-3)] x 50

A2 = ¹/₂(5 + 4) x 50

A2 = 225 J

<h3>Total work done </h3>

W = A1 + A2

W = 20 J + 225 J

W = 245 J

Learn more about work done here: brainly.com/question/8119756

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The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t

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Answer:

The charge is $2.75\times10^{-13}\ C$

Explanation:

Given that,

Distance = 2.5 mm

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Using formula of linear charge density

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$\lambda=\dfrac{Er}{2k}$

Put the value into the formula

$\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}$

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Using formula of charge

$Q=\lambda\timesL$

Put the value into the formula

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