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maks197457 [2]
1 year ago
14

PLEASE HELP a current of 0.5 A flows with a resistance of 50 ohms what is the e.m.f applied NEED HELP ASAP AND BRAINLIEST​

Physics
1 answer:
Vanyuwa [196]1 year ago
6 0

Answer:

25 volts

Explanation:

voltage = current × resistance

v = i \times r \\ v = 0.5 \times 50 \\ v = 25v

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How much heat in joules is required to melt an ice cube with a mass of 18.6 g at 0 °C? The Lf for water is 333 J/g.
EastWind [94]

Answer:

The heat energy required, Q = 6193.8 J

Explanation:

Given,

The mass of ice cube, m = 18.6 g

The heat of fusion of ice, ΔHₓ = 333 J/g

The heat energy of a substance is equal to the product of the mass and heat of fusion of that substance. It is given by the equation,

                               <em> Q = m · ΔHₓ     joules</em>

Substituting the given values in the above equation

                                Q = 18.6 g x 333 J/g

                                    = 6193.8 J

Hence, the heat required to melt the ice cube is, Q = 6193.8 J

5 0
3 years ago
A bare 4 AWG copper conductor Installed horizontally near the bottom or vertically, and within that portion of a concrete foun d
neonofarm [45]

Answer:

20 ft

Explanation:

This is the length where the direct contact could be used as ground electrode

4 0
3 years ago
A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
3 years ago
Skydiver jumps out of a plane she falls down word and very fast P. When she Open Sarah parachute she slows down. What force pull
Thepotemich [5.8K]
There are two forces acting on a parachute with a parachutist: the force of gravity and the air resistance. The force pulling the skydiver to the ground would be Wright force
4 0
2 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

8 0
1 year ago
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