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ivolga24 [154]
2 years ago
14

Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type

of wave is this
Physics
2 answers:
Ghella [55]2 years ago
7 0

Answer:

The waves are longitudinal.

Explanation:

There are two types of waves:

Longitudinal waves : The waves in which the particles of medium vibrates in the same direction a the wave travels. Ex: sound waves  

Transverse waves : The waves in which the particles of medium in the perpendicular direction of the wave propagation. Ex: light waves.

The wave is travelling along the length of the slinky toy. So, the direction of motion of particles is same as the direction of motion of wave.

So, the waves are longitudinal in nature.

GrogVix [38]2 years ago
3 0

Answer:

"Longitudinal wave" is the appropriate answer.

Explanation:

  • Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.
  • A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.
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Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.
Dennis_Churaev [7]

Answer:

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})} (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

I_c=\frac{1}{2}MR^2\\\\I_r=mR^2

Finally, by replacing in (1):

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

hope this helps!!

7 0
3 years ago
A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses
stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

<u>k = 17043.5 N/m = 17.04 KN/m</u>

5 0
3 years ago
If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?
sweet [91]

Answer:

slbohohohohhoohhoohooohhoohho

6 0
2 years ago
Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.
Nataly_w [17]

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

     4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

So Given values of Planck Constant are equivalent in English system and metric system.

7 0
3 years ago
I need help with this
lord [1]

Answer:

0.832

Explanation:

8.320 x 10 to the negative 1st power is 0.832

7 0
3 years ago
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