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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?

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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Answer:

Given that

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Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

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C. Using

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So

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2 years ago

Read 2 more answers

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

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Answer:

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$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

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$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

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3 years ago

A bald eagle is flying to the left with a speed of 34 meters

Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U + at

Here,

U = 34 m/s

a = - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V = 34 + (-8)3

= 34 - 24

V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

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2 years ago

A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push

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Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

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3 years ago

A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the subme

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Answer:

The volume of the submerged part of her body is $0.0622m^{3}$

Explanation:

Let's define the buoyant force acting on a submerged object.

In a submerged object acts a buoyant force which can be calculated as :

$B=$ ρ.V.g

Where ''B'' is the buoyant force

Where ''ρ'' is the density of the fluid

Where ''V'' is the submerged volume of the object

Where ''g'' is the acceleration due to gravity

Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.

We can write :

$W_{girl}=B$ (I)

Where ''W'' is weight

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$B=W_{girl}$

$B=610N$

ρ.V.g = 610N

$1000\frac{kg}{m^{3}}.V.(9.81\frac{m}{s^{2}})=610N$ (II)

The force unit ''N'' (Newton) is defined as

$N=kg.\frac{m}{s^{2}}$

Using this in the equation (II) :

$(9810\frac{N}{m^{3}}).V =610N$

$V=\frac{610N}{9810\frac{N}{m^{3}}}$

$V=0.0622m^{3}$

We find that the volume of the submerged part of her body is $0.0622m^{3}$

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3 years ago