Processes in which matter changes between liquid and gaseous states are vaporization, evaporation, and condensation.
Answer:
1. to relate the type of box material to the temperature of the air within the box
2. Question is incomplete
3. scatterplot
Explanation:
1. The only thing done differently in this experiment is the type of material used in making the boxes, hence the experiment must be about that. Before proceeding to answering this question, we must have this at the back of our minds.
We can gather from the experiment that the boxes are of the same size and were subjected to sunlight for an hour (the same time duration for both). Hence, the temperature of the air inside the box will only be affected by the type of material the box is made of since the boxes have the same size and were subjected to sunlight for the same duration.
From the options provided, the best description for this experiment is; to relate the type of box material to the temperature of the air within the box.
2. The question is incomplete. The value for speed/velocity needed to calculate the average time is missing.
However, the formula needed here is velocity = distance ÷ time
3. There are two variables in this experiment; distance and time
The type of graph that shows two variables on it (of the options provided) is the scatterplot.
Answer:
Please screenshot the bar graph so that i can answer.
Explanation:
<h2>
Answer:</h2>
MT⁻¹
<h2>
Explanation:</h2>
Given equation:
F = Kv --------------(i)
Where;
F = Magnitude of force experienced by the object
v = speed at which the object is moving
K = constant.
To get the dimension of K, follow the following steps:
i. <em>Make K subject of the formula in equation(i)</em>
K = F / v
ii.<em> Get the dimension of force F and speed v</em>
Dimension of force = MLT⁻²
Dimension of velocity = LT⁻¹
iii.<em> Substitute these dimensions into the result of (i) above;</em>
K = MLT⁻² / LT⁻¹
iv. <em>Simplify the result in (iii)</em>
<em>First, the </em><em>L</em><em>s in the numerator and denominator will cancel out</em>
K = MT⁻² / T⁻¹
<em>Next, the </em><em>T</em><em>s will be expressed as follows</em>
K = MT⁻² x T¹ [using the law of indices]
K = MT⁻²⁺¹
K = MT⁻¹
Therefore, the dimension of K is MT⁻¹
