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3 years ago

Where is the best place to put pfds while you are out on your boat?

1 answer:

3 0

The best place to put PFDs while you are out on your boat is at the top deck of your boat. PFDs must be placed at an area where it is easily accessible, especially in times of accidents, without any unnecessary gear or objects covering the PFDs, but it is best to wear a life jacket during the whole duration of the boat trip.

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Two long, straight parallel wires 8.2 cm apart carry currents of equal magnitude I. The parallel wires repel each other with a f

o-na [289]

Answer:

36.22 mA

Explanation:

i1 = I , i2 = I, d = 8.2 cm = 0.082 m

Force per unit length = 3.2 nN/m = 3.2 x 10^-9 N/m

μo = 4 π × 10^-7 Tm/A

The formula for the force per unit length between the two wires is given by

F = μo / 4π x (2 i1 x i2) / d

3.2 x 10^-9 = 10^-7 x 2 x I^2 / 0.082

I = 0.0362 A = 36.22 mA

4 0

3 years ago

Find the resistance of an electric light bulb if a current of 0.08 A flows when the potential difference across the bulb is 120

lora16 [44]

U=RI Ohm's law
then R=U/I
=120/0.08
=2250Ω
hope this helps you

8 0

3 years ago

Is the process of changing from one phase of matter to another a physical change or a chemical change? Explain why. please asap

Andreas93 [3]

The process of changing from one phase of matter to another is a physical matter.

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

8 0

1 year ago

the motion of a particle along a straight line is represented by the position versus time graph above. at which of the labeled p

atroni [7]

Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.

The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

Learn more about acceleration here: brainly.com/question/933224

Learn more: brainly.com/question/25887663

3 0

2 years ago