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Snowcat [4.5K]
1 year ago
5

what happens to the frequency the centripetal force on an object in circular motion is doubled while the object keeps the same r

adius of rotation
Physics
1 answer:
kondaur [170]1 year ago
3 0

The new frequency also double the centripetal force on an object in circular motion is doubled while the object keeps the same radius of rotation.

The mass of the object is m .

The radius of the circle is r.

The new centripetal force F2 = 2F1 .

A force that causes a body to follow a curved path is known as a centripetal force, which derives from the Latin words centrum, "Centre," and petered, "to pursue." It always moves in a direction that is the opposite of the body's motion and in the direction of the fixed point of the path's immediate Centre of curvature. "A force by which bodies are dragged or driven, or in any other manner tend, towards a point as to a Centre," was how Isaac Newton defined it. The centripetal force that causes astronomical orbits in Newtonian physics is provided by gravity.

The scenario in which a body moves uniformly fast along a circular path is one frequent example of centripetal force. Towards the Centre of the object, the centripetal force is directed along the radius and at an angle to the velocity.

Learn more about centripetal force  here:

brainly.com/question/11324711

#SPJ4

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If are spaced closely together on the map,there is a drastic temperature change over the distance
AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

6 0
3 years ago
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"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swin
Bingel [31]
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N. 

<span>Fx = [(233 + 840)/g]*v²/7.5 </span>

<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>

<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>

<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>

<span>233 + 840 = Ti*cos40º </span>

<span>solve for Ti. (This is the answer to the part b) </span>

<span>Horizontally </span>

<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>

<span>Solve for Th </span>

<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>

<span>using v and Ti computed above.</span>
3 0
3 years ago
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

a_c =w^2r = \frac{v^2}{r}

in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

a_c = \frac{8^2}{28.2}\ m/s^2

a_c = 2.27\ m/s^2

The answer is the last option

3 0
3 years ago
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

7 0
3 years ago
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
Wittaler [7]

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

6 0
3 years ago
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