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2 years ago

When an ice cube sits out on a warm day and melts where does the heat go?

Physics

2 answers:

arsen [322]2 years ago

7 0

Here ill explain

There isn't any heat , there is a lack of heat. The sun warms the lack of heat (ice) until its just as warm as everything around it . hoped this helped!

prohojiy [21]2 years ago

7 0

Answer:

When an ice cube sits out on a warm day and melts the heat goes into the ice

Explanation:

When an ice cube sits out on a warm day and melts where does the heat go?

Firstly, temperature is the degree of hotness or coldness of a body, while heat is the transfer of internal energy due to temperature difference.

When ice melts there is no temperature change. The heat needed to change it from its solid state to its liquid state any noticeable change in temperature is called Heat of fusion Q measured in joules.

Q=m $L_{f}$

So the heat goes directly to the ice to change its state from solid to liquid without any change in temperature. Only to break its intermolecular force of attraction.

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Why is it good practice for scientists to repeat trails

tensa zangetsu [6.8K]

It makes the data thet they collect more reliable so if they need the data again, they have already tested it a few times so therefor they know that it is right.

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3 years ago

Atoms are happy (they will not readily react with other elements) when they have a full outside ring of

Len [333]

Answer: TRUE

Explanation:

Atoms are happy when they will not react with other elements while having a full outside ring of electrons because this makes them to be noble.

A stable atom possesses full outside ring of electrons while unstable one does not. So, they are happy also because of stability.

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3 years ago

Solve this physics for me please with steps

Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

$[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\$

$[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] + [\frac{m^{2} }{s^{2} } ]$

$[\frac{m^2}{s^2} ]$

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

$v=\frac{dx}{dt} \\v = 0.75*2*t+5*t$

$(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]$

ii) replacing in the derivated equation.

$v=1.5*(4)+5\\v=11[m/s]$

iii) the average velocity is defined by the expresion v = x/t

$v = \frac{x-x_{0} }{t-t_{0} } \\$

$x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]$

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

$\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} } \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })$

b) One of the applications of the surface tension is the capillarity this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0

3 years ago

Are dimensionless quantities always unitless

timofeeve [1]

Important thing is that all unitless quantity is dimensionless quantity. .A dimensionless physical quantity may have an unit

4 0

2 years ago

Read 2 more answers

The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1

Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})$ (1)

Where:

$F$ - Explosive force, measured in newtons.

$\Delta s$ - Barrel length, measured in meters.

$m$ - Mass of the shell, measured in kilograms.

$v_{o}$ , $v_{f}$ - Initial and final speeds of the shell, measured in meters per second.

If we know that $m = 1250\,kg$ , $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 750\,\frac{m}{s}$ and $\Delta s = 15\,m$ , then the explosive force experienced by the shell inside the barrel is:

$F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}$

$F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}$

$F = 23437500\,N$

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0

2 years ago