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3 years ago

Where would the barycenter of these two bodies be located given their masses?

Physics

1 answer:

n200080 [17]3 years ago

3 0

It's kinda tough, since we don't know the actual numerical locations of the points, only an approximate picture.

The big ball has 11 times the mass of the small ball, so the small ball is 11 times as far from the barycenter as the big ball is.

If any of the points is marked at the actual barycenter, it can only be point-A .

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Now let’s apply Coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of ot

Mnenie [13.5K]

Answer:

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

q₂ = -6.0 nC = -6.0 10⁻⁹ C

x₂ = 4.0 cm = 4.0 10⁻² m

q₃ = 5.0 nC = 5.0 10⁻⁹ C

x3 = 0 m

Charges q1 and q3

r = x₁ -x₃

r = 2.0 10⁻² -0

r = 2.0 10⁻² m

F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

r = r₂ –r₃

r = 4.0 10⁻² - 0 = 4.0 10⁻² m

F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²

F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

F = F₂₃ - F₁₃

F = 16.86 10⁻⁵ - 33.7 10⁻⁵

F = - 16.84 10⁻⁵ N

F = - 1.68 10⁻⁴ N

The negative sign means that it is directed to the left of the x-axis

5 0

3 years ago

The ideal mechanical advantage of a machine reflects the increase or decrease in force there world be without friction, it is al

brilliants [131]

True: the ideal mechanical advantage of a machine is always greater than the actual mechanical advantage because all machines must overcome friction.

Explanation:

For a simple machine, it is possible to calculate two types of mechanical advantage:

1) The Ideal Mechanical Advantage (IMA) is given by

$IMA=\frac{d_e}{d_r}$

where

$d_r$ is the resistance arm

$d_e$ is the effort arm

The IMA gives the mechanical advantage of the machine if there are no friction forces acting on it, and if all the work in input is converted into work in output with no loss of energy

2) The Actual Mechanical Advantage (AMA) is given by

$AMA=\frac{L}{E}$

where

L is the load (the force in output)

E is the effort (the force in input)

The AMA gives the real mechanical advantage of the machine. For an ideal machine,

$AMA=IMA$

Because there is no loss of energy due to friction.

For a real machine instead,

$AMA$

because part of the input energy is converted into thermal energy and other forms of energy due to the presence of friction, so it is "wasted" energy.

Learn more about levers and machines:

brainly.com/question/5352966

#LearnwithBrainly

8 0

3 years ago

The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

chubhunter [2.5K]

Answer:

The angular separation equals $0.35^{o}$