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4 years ago

Evaporation only occurs at the surface of a liquid

Physics

1 answer:

Veseljchak [2.6K]4 years ago

8 0

yes

evaporation starts on the surface

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A roller coaster is released from the top of a track that is 57 m high. Find the roller coaster speed when it reaches ground lev

Art [367]

Hi there!

Assuming the track is frictionless:

$E_i = E_f\\\\PE = KE\\\\mgh = \frac{1}{2}mv^2$

Cancel out the masses and rearrange to solve for velocity:

$gh = \frac{1}{2}v^{2}\\\\v = \sqrt{2gh}\\\\$

Plug in the given height and let g = 9.8 m/s²:

$v = \sqrt{2(9.8)(57)} = \boxed{33.42 m/s}$

7 0

3 years ago

The sugar molecule known as sucrose is a polar molecule. Which statement BEST summarizes how sucrose will dissolve in water?

kirill [66]

I believe the answer is C

Hope this helps :)

8 0

3 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

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3 years ago

A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring tha

Ostrovityanka [42]

Answer:

0.358Kg

Explanation:

The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

0.5Ke^2 = 0.5Mv^2

0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2

Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

plugging into Energy equation

12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

12.56 = 0.5 × 9.2872/ M + 0.0115

12.56 = 4.6436/ M + 0.0115

12.56 ( M + 0.0115 ) = 4.6436

12.56M + 0.1444 = 4.6436

12.56M = 4.6436 - 0.1444

12.56 M = 4.4992

M = 4.4992÷12.56

M = 0.358 Kg

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3 years ago

Football player A has a mass of 210 pounds and is running at a rate of 5.0mi/hr. He collides with player B. Player B has a mass

LuckyWell [14K]

The answer is b.) the momentum before the collision is greater than the momentum after the collision

8 0

3 years ago

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