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3 years ago

Gelatin has a density of 1.27 g/cm³. if you have a blob of gelatin dessert that fills a 2.0 liter bottle, what is its mass?

2 answers:

7 0

There are 1,000 cm³ in 1 liter, 2,000 cm³ in 2 liters.

Each cm³ of gelatin has 1.27 grams of mass in it.

There are 2,000 cm³ in the whole blob, so the total mass is

(2,000 cm³) · (1.27 g/cm³) = 2,540 grams

or 2.540 kilograms .

Verdich [7]3 years ago

6 0

Density = mass / volume so mass = density x volume

2 litres = 2000cm^3

mass = 1.27 x 2000 = 2540g = 2.54 kg

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2 years ago

How many significant digits are in the following measurements?<br> a. 1300 m

Fofino [41]

Answer:

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Explanation:

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3 years ago

A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against

valkas [14]

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies $V_1=12.7dm^3=12.7liters$

And volume is increased by $3.3dm^3$ so $V_2=12.7+3.3=16liters$

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

$W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj$

(b) At reversible process work done is given by $W=nRTln\frac{V_2}{V_1}$

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles $n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125$

So work done $W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J$

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3 years ago

A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th

Juliette [100K]

Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

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3 years ago

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

VARVARA [1.3K]

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

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4 years ago