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coldgirl [10]
3 years ago
7

What happens if balanced forces are applied to a moving object? A. The object stops moving. B. The object moves faster in the sa

me direction. C. The object moves slower in the same direction. D. The object keeps moving at the same speed and direction.
Physics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

The object keeps moving at the same speed and direction.

Explanation:

This is because the balanced forces acting on the object cancel out each other since the are in opposite directions, leaving only the original force acting on the object which is causing its motion.To illustrate, let F₁ be the initial force and F₂ and F₃ be the balanced forces. The net force F = F₁ + F₂ + F₃. Since F₂ and F₃ are balanced, F₂ = -F₃ ⇒ F₂ + F₃ = 0.

So,

F = F₁ + F₂ + F₃

  = F₁ + (F₂ + F₃)

  = F₁ + 0

  = F₁

Which is the original force acting on the object. Thus, the object keeps moving at the same speed and direction

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A circular coil of five turns and a diameter of 30.0 cm is oriented in a vertical plane with its axis perpendicular to the horiz
____ [38]

The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

B(h) = B(coil) = μ₀ NI / 2R

B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3

B(h) = 12.6 μT

Learn more about magnetic field here:

brainly.com/question/14411049

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5 0
2 years ago
How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the
kirza4 [7]

Answer:

Approximately 6.2\; {\rm rpm}, assuming that the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}.

Explanation:

Let \omega denote the required angular velocity of this Ferris wheel. Let m denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

  • Weight of the passenger (downwards), m\, g, and possibly
  • Normal force F_\text{normal} that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is 0- that is, F_\text{normal} = 0.

The net force on this passenger is (m\, g - F_\text{normal}). Hence, when F_\text{normal} = 0, the net force on this passenger would be equal to m\, g.

Passengers on this Ferris wheel are in a centripetal motion of angular velocity \omega around a circle of radius r. Thus, the centripetal acceleration of these passengers would be a = \omega^{2}\, r. The net force on a passenger of mass m would be m\, a = m\, \omega^{2}\, r.

Notice that m\, \omega^{2} \, r = (\text{Net Force}) = m\, g. Solve this equation for \omega, the angular speed of this Ferris wheel. Since g = 9.81\; {\rm m\cdot s^{-2}} and r = 23\; {\rm m}:

\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}.

\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}.

The question is asking for the angular velocity of this Ferris wheel in the unit {\rm rpm}, where 1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s}). Apply unit conversion:

\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}.

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2 years ago
The rock cycle _____.
Reika [66]
B is the answer you need and i honestly got this question on a middle school test

you must be in different area then me

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Answer:

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