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3 years ago

What changes occur at the molecular level when butter melts in a hot pan?

Physics

2 answers:

Gala2k [10]3 years ago

7 0

Answer:B. the motion of the butter molecules increases

Explanation:The melting point of the butter lies between 32°C to 35°C, it has no sharp melting point because it is chemically a mixture with no definite composition.

When the heat is supplied to the butter then its molecules gain the energy and their kinetic energy increases which leads to the change in its phase.

Komok [63]3 years ago

5 0

The Answer is B. The Motion of the butter molecules remains the same

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What type of energy does an object gain as it is lifted at a constant speed?

marshall27 [118]

D. Gravitational potential

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3 years ago

A few weeks later, you attend a rock concert at which the band plays at 120dB. How many times more intense is the rock concert t

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4 years ago

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement

eimsori [14]

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

$a(t)=5+4t-2t^2$

Also, $a=\dfrac{dv}{dt}$

$v=\int\limits {a.dt}$

$v=\int\limits {(5+4t-2t^2).dt}$

$v=5t+2t^2-\dfrac{2}{3}t^3+c$

At t = 0, $v(0)=3\ m/s$ . So, c = 3

$v=5t+2t^2-\dfrac{2}{3}t^3+3$

Also, $v=\dfrac{ds}{dt}$ , s is the position

$s=\int\limits {v.dt}$

$s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}$

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'$

At t = 0, $s(0)=10\ m$ . So, c' = 10

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10$

At t = 4 s

$s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10$

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.

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3 years ago

Suppose there are two identical gas cylinders. One contains the monatomic gas krypton (Kr), and the other contains an equal mass

tensa zangetsu [6.8K]

Answer:

Explanation:

Let equal mass of Ne and Kr be m gm

no of moles of Ne and Kr will be m / 20 and m / 84 ( atomic weight of Ne and Kr is 20 and 84 )

Let the pressure and volume of both the gases be P and V respectively .

The temperature of Ne be T₁ and temperature of Kr be T₂.

For Ne

PV = (m / 20) x R T₁

For Kr

PV = (m / 84) x R T₂

T₁ / T₂ = 84 / 20

We know that

average KE of an atom of mono atomic gas = 3 / 2 x k T

k is boltzmann constant and T is temperature .

KEKr/KENe = T₂ / T₁

= 20 / 84

4 0

3 years ago

Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?

Sholpan [36]

You will have to redefine the machine's boundaries.

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4 years ago

Read 2 more answers