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ivolga24 [154]
3 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]3 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]3 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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A closed loop conductor that forms a circle with a radius of 1 m is located in a uniform but changing magnetic field. If the max
lora16 [44]

Answer:

1.59 Tesla per second

Explanation:

Given that

Radius of the conductor loop, r = 1 m,

Maximum induced emf to the loop, e = 5 V,

We take an assumption, we assume that the rate which the magnetic field is changing is dB / dt.

We then go ahead to apply Farady's law of electromagnetic induction

e = rate of change of magnetic flux

e = dФ /dt

e = A * dB / dt, on substituting we see that

A = πr²

A = π * 1²

A = π, plug in the value for A

5 = 3.14 * dB / dt

dB / dt = 5/3.142

dB / dt = 1.59 Tesla per second

8 0
3 years ago
An aircraft, traveling northward, lands on a runway with a speed of 77 m/s. Once it touches down, it slows to 6.3 m/s over 705 m
KengaRu [80]

Answer:

magnitude

a = 4.18 m/s^2

Direction

Opposite to the velocity of aircraft

Explanation:

As we know that the initial speed of the aircraft when it land on the ground is

v_i = 77 m/s

now the final speed of the aircraft when it covers 705 m on runway is given as

v_f = 6.3 m/s

so here we know that

v_f^2 - v_i^2 = 2 a d

so now we can say that

6.3^2 - 77^2 = 2ad

6.3^2 - 77^2 = 2(a)(705)

a = -4.18 m/s^2

5 0
3 years ago
A box is initially at rest on a horizontal, frictionless table. If a force of 10
lukranit [14]

Answer: 18

Explanation:

i think

8 0
3 years ago
Why just indirect evidence be used to study the structure of atoms?
ArbitrLikvidat [17]
It has to be used because you can not visually see them
6 0
3 years ago
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges
Nostrana [21]

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

8 0
3 years ago
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