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Paladinen [302]
3 years ago
14

The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids

and water is heated at a rate of 100 W for 5 minutes. During this time, the temperature of the system increases by 20oC. What is the mass of the lipid content, to the nearest hundredth of a kg, in this solution?
Physics
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

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tigry1 [53]

Answer:

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B = μ(0) [I - i'] / 2πr

where I = 24, and i' = 0, this means that

B = [12.568*10^-7 (24 - 0) ] / 2 * 3.142 * 0.001

B = 3.016*10^-5 / 0.006284

B = 0.0048 T

b), formula is the same, except for change in values, so we have

I = 24, and

i' = (3y2 - 2y2) / (4y2 - 2y2) * 24

i' = 10, and substituting these in the equation, we have

B = [12.568*10^-7 (24 - 10) ] / 2 * 3.142 * 0.003

B = (12.568*10^-7 * 14) / 0.018852

B = 1.76*10^-5 / 0.018852

B = 0.00093 T

and for the last one, we have

I = 24 and i' = 24 also, so that

B = [12.568*10^-7 (24 - 24) ] / 2 * 3.142 * 0.005

B = 0 / 0.03142

B = 0

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3 years ago
Which scale of measurement measures the magnitude or strength of an earthquake based on seismic waves?
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The answer is Richter
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A charge q is at the point x = 5.0 m , y = 0. Write expressions for the unit vectors you would use in Coulomb's law if you were
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Answer:

A) \^r = \^y

B) \^r = -\^x

C) \^r = -0.14\^x + 0.99\^y

Explanation:

Coulomb's Law is

\vec{F}_{ab} = K\frac{q_1q_2}{r^2}\^r

where r is the distance between the two point charges.

The question clearly asks the unit vector expressions of the force that q exerts on the other charge. So, we do not need to find the force, but only the distance vector, r, between charges, and then we can derive the unit vector pointing only the direction of the force.

A) \vec{r} = \vec{r}_b - \vec{r}_a = (5\^x + \^y) - (5\^x + 0) = \^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{\^y}{1} = \^y

B) \vec{r} = \vec{r}_b - \vec{r}_a = (0 + 0) - (5\^x + 0) = -5\^x\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-5\^x}{5} = -\^x

C) \vec{r} = \vec{r}_b - \vec{r}_a = (4.5\^x + 3.5\^y) - (5\^x + 0) = -0.5\^x + 3.5\^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-0.5\^x + 3.5\^y}{\sqrt{(0.5)^2+(3.5)^2}} = \frac{-0.5\^x + 3.5\^y}{3.53} = -0.14\^x + 0.99\^y

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A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced sli
goblinko [34]

Given Information:  

Wavelength of the red laser = λr = 632.8 nm

Distance between bright fringes due to red laser = yr = 5 mm

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Required Information:  

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Answer:

Wavelength of the laser pointer = λp = ?

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

y = Dλ/d  

y/λ = D/d

Where

λ is the wavelength

y is the distance between bright fringes.

d is the double slit separation distance

D is the distance from the slits to the screen

For the red laser,

yr/λr = D/d

For the laser pointer,

yp/λp = D/d

Equating both equations yields,

yr/λr = yp/λp

Re-arrange for λp

λp = yp*λr/yr

λp =  (5*632.8)/5.14

λp = 615.56 nm

Therefore, the wavelength of the small laser pointer is 615.56 nm.

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