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Amanda [17]
3 years ago
11

An ethylene glycol solution contains 25.4 g of ethylene glycol (C2H6O2) in 89.0 mL of water. (Assume a density of 1.00 g/mL for

water.) You may want to reference (Pages 528 - 538) section 14.6 when completing this problem. Part A Determine the freezing point of the solution.
Physics
1 answer:
Vaselesa [24]3 years ago
8 0

Answer:

-7.44°C

Explanation:

Calculate the molality of the solution. Use the density of the solvent(water) as a conversion factor in order to convert from millilitres of solvent to grams of solvent. Then convert grams into kilograms. Finally, use the molar mass of ethylene glycol as a conversion factor to convert from grams to moles of ethylene glycol.  

m = 25.4 g C2H6O2/89.0 mL solv

   = 4.6321 C10H8O

Compute the freezing-point depression.  

ΔT_f=K_f*m ==> (1.86°C)*(3.9996 m)

       =7.44°C

Compute the freezing point of the solution by subtracting the freezing-point depression to the freezing point of the pure solvent.  

 freezing point =0.0°C-ΔT_f

                        = -7.44°C

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storchak [24]
The figure shown below illustrates a lever being used to lift a weight W.
The applied force is F.

In order to lift the weight, the minimum applied force is
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F = (a/b)W

When the ratio a/b < 1, then the applied force is less than the weight.
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Answer: A
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8 0
3 years ago
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Vector A→, having magnitude 2.5m, pointing 37∘ south of east and vector B→ having magnitude 3.5m, pointing 20∘ north of east are
ahrayia [7]

Answer:

Magnitude of the resultant vector is R = 6.81 m

Explanation:

Given :

Vector A having magnitude of 2.5 m

Vector A having direction 37 degree south of east.

Vector B having magnitude of 3.5 m

Vector B having direction 20 degree north of east.

Therefore, the angle between the two vectors is, θ = 37+20 = 57 degree

So, the resultant of the two vectors are given by

$ R = \sqrt{A^2+B^2+2AB \cos \theta}$

$ R = \sqrt{2.5^2+3.5^2+2(2.5)(3.5) \cos 57}$

$ R = \sqrt{6.25+12.25+17.5 \times 0.54}$

$ R = \sqrt{46.45}$

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3 0
3 years ago
Which two quantities can be used to describe motion? A: Displacement and weight. B: speed and acceleration. C: speed and mass. D
HACTEHA [7]
The answer is B as all the other options contain quantities not related to describing motion
3 0
3 years ago
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with t
Mamont248 [21]

Answer:

a)h_{max}=14536.16 m

b)h = 15687.9 m

c)PD=7.62\% The estimate is low.

Explanation:

a) Using the energy conservation we have:

E_{initial}=E_{final}

we have kinetic energy intially and gravitational potential energy at the maximum height.

\frac{1}{2}mv^{2}=mgh_{max}

h_{max}=\frac{v^{2}}{2g}

h_{max}=\frac{43^{2}}{2*0.0636}

h_{max}=14536.16 m  

b)  We can use the equation of the gravitational force

F=G\frac{mM}{R^{2}}   (1)

We have that:

F = ma    (2)

at the surface G will be:

G=\frac{gR^{2}}{M}

Now the equation of an object at a distance x from the surface.

is:

F=\frac{mgR^{2}}{(R+x)^{2}}

m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}

Using that dv/dt is vdx/dt and integrating in both sides we have:

v_{0}=\sqrt{\frac{2gRh}{R+h}}

h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}

h=15687.9

c) The difference is:

So the percent difference will be:

PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%

PD=7.62\%

The estimate is low.

I hope it helps you!

7 0
3 years ago
Calculate the recoil velocity in the horizontal direction, in meters per second, of a 1.25-kg plunger that directly interacts wi
mart [117]

Answer:

v_1=-8.19\ m/s'

Explanation:

It is given that,

Mass of the plunger, m_1=1.25\ kg

Mass of the bullet, m_2=0.0175\ kg

Initially both plunger and the bullet are at rest, u_1=u_2=0

Final speed of the bullet, v_2=585\ m/s

Let v_1 is the final speed of the plunger. Using the conservation of momentum to find it. The equation is as follows :

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m_1v_1+m_2v_2=0

v_1=-\dfrac{m_2v_2}{m_1}

v_1=-\dfrac{0.0175\times 585}{1.25}

v_1=-8.19\ m/s

So, the recoil velocity of the plunger is 8.19 m/s. Hence, this is the required solution.

7 0
3 years ago
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