Answer: Solution W and Y solution have more solubility than X and Z
Solutions are homogeneous mixtures of two or more components. By uniform mix we mean that its structure and properties are the same in the whole mix. Generally, the component which is present in the largest quantity is known as solvent. Solvent determines the physical condition in which the solution exists. In addition to the solvent, one or more component present in the solution is called solutes. In this unit we will only consider binary solutions (i.e., with two components)
The structure of the solution can be described by expressing its concentration. The latter can either be expressed qualitatively or quantitatively. For example, in qualitatively we can say that the solution is diluted (i.e., relatively small amounts of solubility) or it is concentrated (i.e., relatively rarely sighs). But in real life such details may be very confusing and thus require a quantitative description of the solution. There are several ways that we can quantitatively describe the concentration of solutions. (i) Mass Percentage (W / W): The mass percentage of a component of the solution is defined as: mass of the component = mass of the component in the solution = 100 Total mass of the solution .For example, if by mass A solution is described by 10% glucose in water, it means that 10 grams of glucose dissolved in 90 grams of water, resulting in 100 grams of solution. The concentration described by a large percentage of the population is usually used in industrial chemical applications. For example, the commercial bleaching solution contains 3.62 mass percentages of sodium hypochlorite in water. (ii) Volume Percentage (V / V): Volume Percentage is defined as: Total Volume of Component Volume 100 (component) Volume% of Component
Explanation:
Explanation:
Let f is the frequency of an oscillation and T is the period of the oscillation. There exists an inverse relationship between the frequency and the time period of the oscillation. Mathematically, it is given by :
Also, 
So,
The time taken to complete one oscillation is called the period of the oscillation and the number of oscillation is called the frequency if an oscillation.
Answer:
Option C. 30 m
Explanation:
From the graph given in the question above,
At t = 1 s,
The displacement of the car is 10 m
At t = 4 s
The displacement of the car is 40 m
Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:
Displacement at t = 1 s (d1) = 10 m
Displacement at t= 4 s (d2) = 40
Displacement between t = 1 and t = 4 (ΔD) =?
ΔD = d2 – d1
ΔD = 40 – 10
ΔD = 30 m.
Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.
Answer:
The speed change during the 45-minute trip is 20[mph]
Explanation:
When we see the speed at the 45 minutes this is 20 [mph] and at the 0 minutes the speed is 0 [mph].
Therefore the change is (20 - 0) = 20 [mph]
In the attached image we can see the different figures. In fig 1 we can see the bicycle's speed after 10 minutes when the speed becames constant.
In the fig. 2 we can find the graph when the biker stopped at 30 minutes and took a 15-minute break.
Figures 3 and 4, show the differences when a horizontal line is traced on a position vs time graph, and when the horizontal line is traced in a speed vs time graph.
For fig 3 we can conclude that the body is not moving therefore there is no velocity or acceleration. And for the fig 4, we can realize that the area under the horizontal line represents a displacement during the respective interval of time.
given:
mass = 5000 kg
u (initial velocity) = 0 m/s
v (final velocity) = 70 m/s
time taken to change velocity = 3600 s
acceleration = v - u / t
a = 70 - 0 / 3600
a = 70 / 3600
a = 0.0194 m/s2 (approx)
given: mass = 5000 kg
acceleration = 0.0194 (found in 1st part)
force = mass * acceleration
f = ma
f = 5000*0.0194 = 97 N
therefore the acceleration will be 0.0194 m/s2
and the force involved in acceleration will be 97 N
