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Natasha2012 [34]
3 years ago
14

Which one of the following scenarios accurately describes a condition in which resonance can occur? A. A vibrating tuning fork i

s struck and begins to vibrate as the object used to strike it is placed away from the tuning fork. B. A person is holding a tuning fork on one end of a football field and another person is holding a separate tuning fork on the opposite end of the field. C. A column of air has a height equal to 1⁄8 of the wavelength of the sound waves produced by a tuning fork vibrating over the column of air in an open pipe that's partially immersed in water. D. A pipe's length is equal to 1⁄2 of the wavelength of the sound waves produced by a tuning fork vibrating over one end of the pipe that's open to the air at both ends.
Physics
1 answer:
Digiron [165]3 years ago
5 0
I believe the answer is A
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What is 3,278,000 as scientific notation?
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4 0
3 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor
Serggg [28]

Answer:

a = 1 m/s²  and

Explanation:

The first two parts can be seen in attachment

We use Newton's second law on each axis

Y axis

      Ty - W = 0        

      Ty = w

X axis

     Tx = m a

With trigonometry we find the components of tension

    Sin θ = Ty / T

    Ty = T sin θ

    Cos θ = Tx / T

    Tx = T cos θ

We calculate the acceleration with kinematics

   Vf = Vo + a t

   a = (Vf -Vo) / t

   a = (20 -10) / 10

   a = 1 m/s²

We substitute in Newton's equations

     

  T Sin θ = mg

  T cos θ = ma

We divide the two equations

  Tan θ = g / a

  θ = tan⁻¹ (g / a)

  θ = tan⁻¹ (9.8 / 1)

  θ = 84º

We see that in the expression of the angle the mass does not appear therefore you should not change the angle

4 0
3 years ago
An unstretched spring has a length of 0.30 m. When the spring is stretched to a total length of 0.60 m, it supports traveling wa
IceJOKER [234]

Explanation:

Below is an attachment containing the solution

7 0
3 years ago
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