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zlopas [31]
3 years ago
5

A boat moves through the water with two forces acting on it. One is a 2.51×103 N forward push by the motor, and the other is a 1

.88×103 N resistive force due to the water. What is the acceleration of the 1384.1 kg boat? Answer in units of m/s 2
Physics
1 answer:
telo118 [61]3 years ago
7 0

Answer:

Acceleration a=0.455m/sec^2

Explanation:

We have given that the force by the motor F=2.51\times 10^3N=2510N

Resistive force by the water F=1.88\times 10^3N=1880N

So net force F_{NET}=2510-1880=630N

Mass of the boat = 1384.1 kg

According to newton's second law F=Ma, here M is mass and a is acceleration

So 630=1384.1\times a

a=0.455m/sec^2

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A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘
hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

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x + y = 19.12 g..(1)

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Freezing point of the solution,T_f = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC

\Delta T_f=i\times K_f\times m

Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}

5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})

On solving we get:

\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of Ca(NO_3)_2 in the mixture:

\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%

The mass percentage of calcium nitrate is 31.23%.

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