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Assoli18 [71]
3 years ago
13

A glass of water sitting in direct sunlight evaporates over time. Explain this phase change in terms of the types of heat transf

er that occur.
Physics
2 answers:
olga_2 [115]3 years ago
6 0
Well over time the temperature of the water increase causeing its moleclues to move alot which causes it's state change to a gas this is who we have rain.
AlexFokin [52]3 years ago
4 0

Answer:

Heat transfer occurs by radiation

Explanation:

There are three ways by which heat can be transferred from one body to another.

Conduction occurs when the two bodies are in contact. Heat transfer occurs due to vibration and collision of molecules.

Heat transfer via convection occurs when there is bulk motion of fluid.

Heat transfer via radiation can occur through space.

A glass of water kept in sunlight evaporates as the heat transfers to water via radiation. The temperature of the water rises and it starts evaporating.

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If i ask wat r the application of simple machine some of u will say-----> Simple machines that are widely used include the wh
TiliK225 [7]

Answer:

DID NOT

UNDER STAND

Explan it a littlebit

3 0
3 years ago
Any child is pushing a shopping cart at a speed of 1.5 m/s.how long will it take this child to push the cart down the aisle with
NARA [144]
1.5 m/s is the velocity. 9.3 m is the length of aisle, over which Distance will be covered. Time is demanded in which the child will move the cart over the aisle with 1.5 m/s. v=S/t and, t=S/v Put values, t=9.3/1.5=6.2 s
7 0
3 years ago
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

And we have 60^\circ F in t=30 s, So:

T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

8 0
3 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
An object starts at rest. Its acceleration over 30 seconds is shown in the graph below:
ddd [48]

Answer:

The instantaneous speed of the object after the first five seconds is 12.5 m/s.

(C) is correct option.

Explanation:

Given that,

An object starts at rest. Its acceleration over 30 seconds.

We need to calculate the instantaneous speed of the object after the first five seconds

We know that,

Area under the acceleration -time graph gives speed.

According to figure,

speed = area\ of\ tringle

speed=\dfrac{1}{2}\times b\times h

speed =\dfrac{1}{2}\times5\times5

speed-12.5\ m/s

Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.

6 0
3 years ago
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