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iren [92.7K]
2 years ago
8

2. Do you think the density of the ice affected the melting rate of the ice, or do you think adding the objects affected the mel

ting rates?
Physics
1 answer:
Setler [38]2 years ago
3 0

The density of ice does not affect its melting rate. Adding objects will affect the melting rate.

  • A physical process called melting or fusing causes a substance to change its phase from a solid to a liquid. This happens when the solid's internal energy rises, usually as a result of heat or pressure being applied, which raises the substance's temperature to the melting point.
  • The term "density" refers to an extensive quality, which means that it is independent of the substance's concentration. Every substance in the world demonstrates its distinctive density. Since it does not fluctuate, it would not affect the rate of melting. The addition of the objects could speed up the process, though, as each one generates heat that could act as the mediating force for the melting process.

To learn more about density, visit :

brainly.com/question/15164682

#SPJ9

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A soap bubble has a diameter of 4mm .Cakculate the pressure inside it if the atmospheric pressure outside is 10^5Nm^2 . The surf
Kipish [7]
Look it up it will give the right anwser
3 0
3 years ago
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The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr
Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

I(t)=0.88e^{-t/6\times3600s}

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

q=\int\limits {I} \, dt

Here the limits of integration is from 0 to infinite. So,

q=\int\limits {0.88e^{-t/6\times3600s}}\, dt

q=0.88\times(-6\times3600)(0-1)

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}

N = 1.2 x 10²³

8 0
3 years ago
A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².​
Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2}  } =0.0001[m^{2} ]

Now replacing:

P=100/0.0001\\P=1000000[Pa]

3 0
2 years ago
Energy slowly leaks outward through the diffusion of photons that repeatedly bounce off ions and electrons
stepan [7]

Energy slowly leaks outward through the radiative diffusion of photons that repeatedly bounce off ions and electrons.

<h3>What is radiative diffusion?</h3>

A radiation zone is a layer of a star's core where energy is mostly carried toward the outside by radiative diffusion and thermal conduction rather than convection.

As photons, energy passes through the radiation zone as electromagnetic radiation.

The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.

Hence,radiative diffusion is correct answer.

To learn more about radiative diffusion refer:

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7 0
2 years ago
If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done
ioda

Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J

Therefore, the work done by the lawnmower is 236.14 J.

6 0
3 years ago
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