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3 years ago

Physical science help

Physics

2 answers:

mojhsa [17]3 years ago

4 0

The answer is pulley

VMariaS [17]3 years ago

3 0

The flagpole used a PULLEY at the top.

The cutting edge of the axe is a WEDGE.

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Pls help me... taking a quiz.

saw5 [17]

Guess I recommend doing that

7 0

3 years ago

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.

Total mass of mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

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%.

5 0

3 years ago

explain why when someone uses his thumb to push a pin into a block of wood the pressure on the wood is greater than the pressure

Charra [1.4K]

Answer:

the smaller the area the bigger the pressure

8 0

3 years ago

Read 2 more answers

A very thin circular disk of radius R = 20.00 cm has charge Q = 30.00 mC uniformly distributed on its surface. The disk rotates

lutik1710 [3]

Answer:

$B= 7.5*10^{-15}T$

Explanation:

The magnetic field strenght on the z-axis at a distance d from the center is,

$B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}$

Our values are:

$R=20cm\\Q=30mc\\w=5rad/s\\d=2*10^3cm=20m$

Replacing,

$B= \frac{(4\pi*10^{}-7)(30*10^{-3})(5)(0.2)^2}{8\pi(20)}$

$B= 7.5*10^{-15}T$

3 0

3 years ago

What do you think happened to the energy you transferred to the notebook when you pushed it?

xeze [42]

Answer:

It got transferred to kinetic energy

Explanation:

8 0

3 years ago

Read 2 more answers