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2 years ago

What is an example of vaporization?

Physics

2 answers:

Alenkasestr [34]2 years ago

7 0

Answer:

just search it up you'll get ur answer

Phoenix [80]2 years ago

5 0

Boiling water is what I would put

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numerical question : what is the required heat to raise the temperature of 2 kg parrafin by 10 Celsius if 44000 joules is requir

kirill115 [55]

Answer:

The heat energy required is, E = 2200 J

Explanation:

Given,

The mass of paraffin, m = 2 Kg

The energy required to raise the temperature of the paraffin by 200° C = 44000 J

Then the heat energy required to raise the temperature of the paraffin by 10° C is given by,

Since 44000 J raises temperature by 200° C, then

E = 44000 J / 20

= 2200 J

Hence, the energy required to raise the temperature of the paraffin by 10° C is, E = 2200 J

8 0

2 years ago

a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of wh

pychu [463]

Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!

7 0

2 years ago

How old was the theory of plate tectonicsâthe defining theory of all earth scienceâwhen man first walked on the moon?

solong [7]

What is your question?

7 0

2 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

2 years ago

How do reflection and refraction of waves affect communication?

romanna [79]

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

5 0

2 years ago

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