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Ket [755]
3 years ago
11

Assuming the density of the wine to be 1.0 g/ml, what caloric content does the alcohol (ethanol) in a 6-oz glass of wine (177 ml

) have? express your answer using two significant figures.
Physics
2 answers:
posledela3 years ago
8 0

Answer:

The answer is 133.06 kcal.

Explanation:

having the data:

mw=mass of wine=177 g

me=mass of ethanol=(10.6/100)*177=18.76 g

ne=moles of ethanol=0.408 moles

qe=caloric content=1367*((0.106*177)/(46.07))=556.7 kJ=133.06 kcal

viktelen [127]3 years ago
4 0

mass = 177 x 1.0 g/mL= 177 g mass ethanol = 177 x 10.6/100 = 18.8 g moles = 18.8 g/46.069 g/mol = 0.408 1367 kJ/mol x 0.408 mol= 557.9 kJ

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Bits of paper are attracted to an electrified comb or rod, even though they have nonet charge. How is this possible?
SashulF [63]

Explanation:

An electrified comb is charged comb ( let say by running it through the hair) and when it is brought in the proximity of  pieces of paper, the pieces tend to cling to it. This happens because the charged comb induces an opposite charge in the paper pieces and as opposite charges attract each other, the pieces are clinged.

7 0
3 years ago
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

3 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

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by relation for electric field we have following relation

E = \frac{kq}{x}^2

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FROM FIGURE

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\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

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6 0
3 years ago
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Ann [662]

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6 0
3 years ago
A large crate of mass m is placed on the back of a truck but not tied down. As the truck accelerates forward with an acceleratio
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3 years ago
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