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stiks02 [169]
4 years ago
5

50ml of water is added to a certain amount of solution with a concentration of 0.08g/ml. The final concentration is 0.065 g/ml.

What is the initial volume of water in the solution before the addition was made?
Physics
1 answer:
vladimir1956 [14]4 years ago
7 0

Answer:

<em>216.7 ml</em>

Explanation:

<u>Concentration</u>

The concentration of a substance is defined as the quantity of solute existent in a given quantity of solution. Water is usually a liquid to add to the solute and produce the substance with the required concentration, often expressed in %, gr/ml or any other similar ratio. If the solute is expressed in units of mass and the solution is expressed in units of volume, then the concentration is known as mass concentration or density and is computed as

\displaystyle C=\frac{m_s}{V_t}

Where m_s is the mass of solute and V_t is the total volume of the solution

We know at first  

C_1=0.08\ gr/ml

\displaystyle \frac{m_s}{V_{t1}}=0.08

Operating

m_s=0.08V_{t1}

When we add 50 ml of water and keep the solute unchanged, we have

\displaystyle \frac{m_s}{V_{t1}+50}=0.065

Or equivalently

m_s=0.065(V_{t1}+50)

Equating both masses

0.08V_{t1}=0.065(V_{t1}+50)

0.08V_{t1}=0.065V_{t1}+3.25

Rearranging and simplifying

0.015V_{t1}=3.25

Solving

V_{t1}=216.7\ ml

The mass of the solute is

m_s=0.08V_{t1}=17.3\ gr

We don't know the density of the solute, so we cannot compute its volume which is part of the solution. If we neglect this volume because the concentration is small enough, we can say there were 216.7 ml of water in the initial solution.

If we estimate the density of the solute close to 1 gr/ml, the volume is 17.3 ml, and the initial volume of water in the solution is 216.7-17-3=199.4 ml

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Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force,  a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as:  \overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\

In respect to the magnetic field; \overrightarrow{B}=(0.027 i - 0.15 j) [T]

The magnetic force can be written as;

\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]

Bear in mind q =-1.6021x10^{-19} [C]  

thus,

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.  

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Answer:

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Explanation:

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Answer:

The asteroid's acceleration at this point is 2.71\ m/s^2

Explanation:

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x=6.5t-2.3t^3

The velocity of asteroid is given by :

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

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6.5-6.9t^2=0\\\\t=0.971\ s                      

Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t                        

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a=-13.8\times 0.197\\\\a=-2.71\ m/s^2

So, the asteroid's acceleration at this point is 2.71\ m/s^2 and it is decelerating.

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3 years ago
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To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS. 

~ x greater than or equal to -18

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3 years ago
A car starts from rest and acquires a velocity of 50m/s in 3secs. Calculate i) acceleration ii) distance covered.
mafiozo [28]

Answer: 75.02 m

Explanation:

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v = 50 m/s

t = 3 s

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= 0 × 3 + 1/2 × 16.67 × 3 × 3

= <u>75.02 m</u>

Hope this helps...

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3 years ago
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