I think the North Pole
If I understand this correctly
B. Purchase a small plastic container and mark 1-ounce increments on the outside to determine volume. Pour 5 ounces of water into the container, and place in the freezer for 8 hours. Compare the frozen or ending volume with the liquid or beginning volume.
<h3>How much water expands when frozen?</h3>
Ice is less denser than the liquid form. Water is the only known non-metallic substance that expands when it freezes because it is the unique property of water. Water density decreases and it expands approximately about 9% by volume. For calculating the expansion of water, plastic container is the best option. We know that water expands when the water freezes because it is a unique property of water which allows the survival of aquatic organisms.
So we can conclude that option B is the right answer.
Learn more about water here: brainly.com/question/1313076
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Answer:
2.85 s .
Explanation:
y(t) = y(0) + v₀t + 1/2 gt²
y(t) is vertical displacement , y(0) is initial position , v₀ is initial velocity and t is time required to make vertical displacement and g is acceleration due to gravity.
Here y(0) is zero , v₀ = 14 m/s , g = 9.8 m s⁻² , y(t ) = 0 , as the pumpkin after time t comes back to its initial position, that is ground .
We shall take v₀ as negative as it is in upward direction and g as positive as it acts in downward direction
Put the values in the equation above,
0 = 0 - 14t + 1/2 x 9.8 t²
14 t = 1/2 x 9.8 t²
t = 28 / 9.8
t = 2.85 s .
Answer:
By a factor of 1/4.
Explanation:
The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,
in which 
, 
represent the change in momentum and the time taken for that change.
If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by 
, the following manipulation confirms the answer to this question.
![\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5C%5C%5Csmall%20F_1%20%26%3D%5Csmall%20%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B4%5CDelta%20t%7D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%5Cbigg%5B%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B%5CDelta%20t%7D%5Cbigg%5D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7DF%5Cend%7Baligned%7D)
Here 
is the force that was applied to the object previously.
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It would be autotroph and hetrotroph
