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shepuryov [24]
3 years ago
11

Penny has two dogs: Fluffy (left) and Butch (right).

Physics
2 answers:
Monica [59]3 years ago
3 0
I think it would be D

They would both move in the same way
Kipish [7]3 years ago
3 0
None knows how dogs looks like. One may be 4kg - another 30kg
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A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a tube of length
lora16 [44]

Answer:

T = 4.42 10⁴ N

Explanation:

this is a problem of standing waves, let's start with the open tube, to calculate the wavelength

        λ = 4L / n                 n = 1, 3, 5, ...    (2n-1)

How the third resonance is excited

       m = 3

       L = 192 cm = 1.92 m

       λ = 4 1.92 / 3

       λ = 2.56 m

As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio

      v = λ f

      f = v / λ

      f = 343 / 2.56

      f = 133.98 Hz

       

Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m

The expression for standing waves on a string is

           λ = 2L / n

           λ = 2 0.37 / 2

           λ = 0.37 m

The speed of the wave is

          v = λ f

As we have some resonance processes between the string and the tube the frequency is the same

          v = 0.37 133.98

          v = 49.57 m / s

Let's use the relationship of the speed of the wave with the properties of the string

              v = √ T /μ

              T = v² μ

              T = 49.57²   18

              T = 4.42 10⁴ N

4 0
3 years ago
Read 2 more answers
Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed
likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let t represents the time in seconds and v the speed in meters-per-second.

For 0 < x \le 1:

  • Initial value of v: \rm 2\;m\cdot s^{-1} at t = 0; Hence the point on the segment: (0, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 2\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1.

Similarly, for 1 < x \le 2:

  • Initial value of v is the same as the final value of v in the previous equation at t = 1: t = 2t + 2 = 4; Hence the point on the segment: (1, 4).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 1\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2.

For 2 < x \le 3:

  • Initial value of v is the same as the final value of v in the previous equation at t = 2: t = t + 3 = 5; Hence the point on the segment: (2, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. There's no acceleration. In other words, the velocity is constant.
  • Find the equation of this segment in slope-point form: v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3.

For 3 < x \le 4:

  • Initial value of v is the same as the final value of v in the previous equation at t = 3: t = 5; Hence the point on the segment: (3, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm -3\; m\cdot s^{-2}. In other words, the velocity is decreasing.
  • Find the equation of this segment in slope-point form: v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4.

For 4 < x \le 5:

  • Initial value of v is the same as the final value of v in the previous equation at t = 4: t = -3t + 14; Hence the point on the segment: (4, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. In other words, the velocity is once again constant.
  • Find the equation of this segment in slope-point form: v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5.

t = \rm 3.49\;s is in the interval 3 < x \le 4. Apply the equation for that interval: v = -3t +14 = \rm 3.53\; m \cdot s^{-1}.

4 0
3 years ago
A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c
Marizza181 [45]

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

3 0
3 years ago
A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is
iris [78.8K]
(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the  coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.

(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
\Phi = BA \cos \theta
where \theta is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case \theta=90^{\circ} and so the cosine is zero, therefore the net flux is zero.
5 0
3 years ago
Are any rocks that form from another rock as a result of change in blank or blank
charle [14.2K]
The correct answer is rock cycle
3 0
4 years ago
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