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WARRIOR [948]
3 years ago
7

How does the thermal energy of a large container of water compare to a small container of water at the same temperature

Physics
2 answers:
alex41 [277]3 years ago
6 0

The large container warms slower than the smaller container because of the amount of space . Both can warm to the same temperature tho just one takes longer than the other . Can I please be marked as brainliest! Have a great day !

allochka39001 [22]3 years ago
4 0

The larger container warms up slower than the small one

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A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th
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Answer: V_{f}=2.96m/s    

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight w of the block has an x-component and y-component:

w_{x}=wsin(\theta)    (1)

w_{y}=wcos(\theta)    (2)

As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

N_{y}=Ncos(\theta)    (4)

In addition, we know N=w, then \sum F_{y}=0

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\sum F_{x}=m.a

m.a=w_{x}    (5)

Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

a=5.88m/{s}^{2}    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

d=\frac{1}{2}a{t}^{2}   (10)

We already know the value of  d and calculated a, we have to find t:

t=\sqrt{\frac{2d}{a}}   (11)

t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

V_{f}=a.t   (14)

V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

Finally we get the value of the Final Velocity of the block:

V_{f}=2.96m/s    

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