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Andrei [34K]
3 years ago
14

Your science class decides to conduct an experiment to learn more about ice. You set one ice cube on the counter at room tempera

ture, and another, same-sized ice cube in the class refrigerator. You predict that since it is warmer on the counter at room temperature, the ice cube on the counter will melt faster. What result would confirm this prediction? A)When the counter ice cube is melted, the refrigerator ice cube is not finished melting. B)When the counter ice cube is melted, the refrigerator ice cube is also finished melting. C)When the counter ice cube is not quite melted, the refrigerator ice cube is finished melting. D)When the counter ice cube is not quite melted, the refrigerator ice cube is not finished melting.
Physics
2 answers:
ipn [44]3 years ago
7 0
This would be A. If the counter cube is finished while the refrigerator isn't, that confirms the hypothesis.
anzhelika [568]3 years ago
7 0

Answer:

A) When the counter ice is melted, the refrigerator ice cube is not finished melting.

Explanation:

Considering the varying temperatures, the temperature on the counter at room temperature is higher than that in the refrigerator and as such, the counter ice cube will melt faster than the refrigerator ice cube.

The refrigerator ice cube will also melt, but at a very slow rate compared to the counter ice cube, therefore, when the counter cube is completely melted, the refrigerator ice cube is not finished melting.

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Explanation:

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6 0
3 years ago
Using simple language, explain how your demonstration shows the effect of distance on electric forces. Use the terms electric ch
Eddi Din [679]

Explanation:

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3 years ago
A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one
Fittoniya [83]
I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)
8 0
3 years ago
Read 2 more answers
What is the acceleration of the box?
worty [1.4K]

Answer:

a=4,32m/s^2

Explanation:

Fnet = F1 - F2

= 12-1.2

= 10.8N

m=2.5kg

Fnet =ma

10.8=2.5a then divide both sides by 2.5 to get acceleration

8 0
3 years ago
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