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3 years ago

Write a short story that describes haw matter changes from one physical state to another

Physics

1 answer:

marin [14]3 years ago

3 0

I'll assume this is can or is fictional prose.

Two Icecubes sat in a freezer in the kitchen. One was so bold to claim he could survive walking across the stove. As he made his way to the stove, he noticed he was leaving a wet trail, "no matter" he thought and continued his path. He touched the stove and in an instant was no longer an ice cube. He was a puddle, but not for long. He then started to float and became steam.

Thus he went from solid to liquid to gas.

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If the same quantity of heat is added to both a 2-liter and a 4-liter container of water, the temperature change for water in th

VikaD [51]

Answer:

brainly.com/question/10966794

Explanation:

3 0

2 years ago

PLEASE PLEASE HELP!

vichka [17]

Well, the relationship between the net force and mass and acceleration of an object are directly related, as per the equation - Fnet = ma.

Thus the solution is A. As the net force of an object decreases, the object's acceleration also decreases, mass is kept constant.

5 0

3 years ago

The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s) + 3I2 (s) → 2Al3+ (aq) + 6I− (aq) The emf

Triss [41]

Answer: +2.10V

Explanation:

$2Al(s)+3I_2(s)\rightarrow 2Al^{3+}(aq)+6I^-(aq)$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log K$

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log [Al^{3+}]^2\times [I^-]^6$

where,

$E^o_{cell}$ = standard emf for the cell = +2.20 V

n = number of electrons in oxidation-reduction reaction = 6

$E_{cell}$ = emf of the cell = ?

$[Al^{3+}]$ = concentration = $5.0\times 10^{-3}M$

$[I}^{-}]$ = concentration = $0.10M$

Now put all the given values in the above equation, we get:

$E_{cell}=+2.20-\frac{0.059}{6}\log [5.0\times 10^{-3}]^2\times [0.10]^6$

$E_{cell}=2.10V$

The standard emf for the cell using the overall cell reaction below is +2.10 V

5 0

3 years ago

HELP PLEASE!!! a= ? v= r=

ratelena [41]

Answer:

$a=\frac{v^2}{r}$

$v=\sqrt{a*r}$

$r=\frac{v^2}{a}$

Explanation:

This is the formula for centripetal acceleration in terms of the tangential velocity (v) and the radius of the circular motion (r). The expression for the acceleration is already given, so simply type it as shown:

$a=\frac{v^2}{r}$

For the velocity (v) multiply by "r" both sides and then use the square root to solve for v:

$a*r=v^2\\v=\sqrt{a*r}$

For the radius multiply both sides by r and then divide both sides by the acceleration (a) in order to isolate r completely:

$a*r=v^2\\r=\frac{v^2}{a}$

4 0

2 years ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

valina [46]

Answer:

Explanation:

It is given that,

Mass of lump, m₁ = 0.05 kg

Initial speed of lump, u₁ = 12 m/s

Mass of the cart, m₂ = 0.15 kg

Initial speed of the cart, u₂ = 0

The lump of clay sticks to the cart as it is a case of inelastic collision. Let v is the speed of the cart and the clay after the collision. As the momentum is conserved in inelastic collision. So,

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

v = 3 m/s

So, the speed of the cart and the clay after the collision is 3 m/s. Hence, this is the required solution.

3 0

3 years ago