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prohojiy [21]
3 years ago
14

A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

disk is 5kg and its radius is 2 m. What is the linear speed, in m/s, after the mass drops by 0.3 m?
a. 0.5
b. 1.5
c. 2
d. 3
e.5
Physics
1 answer:
Paha777 [63]3 years ago
8 0

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

E_i =E_f

so:

Mgh = \frac{1}{2}IW^2 +\frac{1}{2}MV^2

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = \frac{1}{2}MR^2

I = \frac{1}{2}(5kg)(2m)^2

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = \frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2

Replacing the data:

(5kg)(9.8)(0.3m) = \frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2

solving for V:

(5kg)(9.8)(0.3m) = V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))

V = 2 m/s

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A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr
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Answer:

388 cm^3

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

pV=const.

which can also be rewritten as

p_1 V_1 = p_2 V_2

In our case, we have:

p_1 = 75.9 kPa is the initial pressure

V_1 = 639 cm^3 is the initial volume

p_2 = 125 kPa is the final pressure

Solving for V2, we find the final volume:

v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3

7 0
3 years ago
A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density
Yuliya22 [10]

The solution of Sulfuric Acid (H2SO4) has the following mole fractions:

  • mole fraction (H2SO4)= 0.034
  • mole fraction (H2O)= 0.966

To solve this problem the formula and the procedure that we have to use is:

  • n = m / MW
  • = ∑ AWT
  • mole fraction = moles of A component / total moles of solution
  • ρ = m /v

Where:

  • m = mass
  • n = moles
  • MW = molecular weight
  • AWT = atomic weight
  • ρ = density
  • v = volume

Information about the problem:

  • m solute (H2SO4) = 17.75 g
  • v(solution) = 100 ml
  • ρ (solution)= 1.094 g/ml
  • AWT (H)= 1 g/mol
  • AWT (S) = 32 g/mol
  • AWT (O)= 16 g/mol
  • mole fraction(H2SO4) = ?
  • mole fraction(H2O) = ?

We calculate the moles of the H2SO4 and of the H2O from the Pm:

MW = ∑ AWT

MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4

MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)

MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol

MW (H2SO4)=  98 g/mol

MW (H2O)= AWT (H) * 2 + AWT (O)

MW (H2O)= (1 g/mol * 2) + (16 g/mol)

MW (H2O)= 2 g/mol + 16 g/mol

MW (H2O)=  18 g/mol

Having the Pm we calculate the moles of H2SO4:

n = m / MW

n(H2SO4) = m(H2SO4) / MW (H2SO4)

n(H2SO4) = 17.75 g / 98 g/mol

n(H2SO4) = 0.1811 mol

With the density and the volume of the solution we get the mass:

ρ(solution)= m(solution) /v(solution)

m(solution) = v(solution) * ρ(solution)

m(solution) = 100 ml * 1.094 g/ml

m(solution) = 109.4 g

Having the mass of the solution we calculate the mass of the water in the solution:

m(H2O) = m(solution) - m solute (H2SO4)

m(H2O) = 109.4 g - 17.75 g

m(H2O) = 91.65 g

We calculate the moles of H2O:

n = m / MW

n(H2O) = m(H2O) / MW (H2O)

n(H2O) = 91.65 g / 18 g/mol

n(H2O) = 5.092  mol

We calculate the total moles of solution:

total moles of solution = n(H2SO4) + n(H2O)

total moles of solution = 0.1811 mol + 5.092  mol

total moles of solution = 5.2731 mol

With the moles of solution we can calculate the mole fraction of each component:

mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution

mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol

mole fraction (H2SO4)= 0.034

mole fraction (H2O)= moles of (H2O) / total moles of solution

mole fraction (H2O)= 5.092  mol / 5.2731 mol

mole fraction (H2O)= 0.966

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

  • Solvent
  • Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

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Answer:

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<h3>You can see that dull surfaces are good absorbers and emitters of infrared radiation. Shiny surfaces are poor absorbers and emitters (but they are good reflectors of infrared radiation</h3>
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