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Marianna [84]
2 years ago
11

The same amount of substance was added to four beakers of water. The treatments were placed in the chart.

Physics
1 answer:
vfiekz [6]2 years ago
5 0

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:

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Which of the following would cause the force of kinetic friction to increase?
MrRa [10]

Answer:

c

Explanation:

4 0
3 years ago
A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking
Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

5 0
3 years ago
In tenis does your knees have to be slightly flexed for both the forehand and backhand?
Elena L [17]

Answer:

yes

Explanation:

8 0
3 years ago
Felipe drives his car at a velocity of 28 m/s. He applies the brake, which slows the vehicle down at a rate of 6.4 m/s2 and caus
zysi [14]
So after ~4.4 seconds, car will stop.

6 0
3 years ago
Read 2 more answers
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
3 years ago
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