<h3>Option D) The red liquid has a smaller specific heat</h3>
When a substance has smaller specific heat it needs less heat to shows changes in it hence making the option D correct.
Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).
Answer:1 because
Explanation: it’s pointing to the earth and gravity
Pulls things down to earth
Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change
Yes heating water allows it to dissolve more Sugars because the molecular distance increases and this distance can be covered by more sugar. In the given question, The independent variable would be the temperature of water.
Since to whatever temperature the water boils at the boiling temperature of does not change remains hundred degree. Rest all the variables can vary the weight of the amount of sugar with the variable in the temperature of Boiling of water to remain constant.
