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Novosadov [1.4K]
3 years ago
9

When you drop a strong magnet through the center of a copper pipe, what happens to the copper piper and the magnet?

Physics
1 answer:
Sedbober [7]3 years ago
8 0
Copper is; unlike iron and steel; not ferromagnetic, but diamagnetic. This means that induced magnetic fields in copper will counter the applied force.

When you drop a strong magnet through a copper pipe, the moving magnetic field will induce currents (Lenz’ Law). These currents will now induce their own magnetic field. This magnetic field counters the falling magnetic.

Result: the magnet will fall way slower than if it was falling through a plastic pipe.
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When you jump straight up as high as you can, what is the order of magnitude of the maximum recoil speed that you give to the Ea
Klio2033 [76]

Answer:

5.66 × 10⁻²³ m/s

Explanation:

If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.

Since initial momentum = final momentum,

mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.

My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?

So mv₁ + M × 0 = m × 0 + MV₂

mv₁ = MV₂

V₂ = mv₁/M =  54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s

5 0
3 years ago
¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?
Zina [86]

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

distancia=\frac{1 segundo*90 km}{3600 segundos}

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

distancia=\frac{0.025 km*1,000 metros}{1 km}

distancia= 25 metros

<u><em>La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h</em></u>

6 0
3 years ago
Humpback whales are known to produce a collection of elaborate and repeating sounds with frequencies starting at 20 Hz. The soun
jasenka [17]

Answer:

70 m.

Explanation:

Given,

Frequency, f = 20 HZ

speed of sound, v = 1400 m/s

wavelength of the waves = ?

we know,

v = f λ

\lambda= \dfrac{v}{f}

\lambda= \dfrac{1400}{20}

\lambda=70\ m

Hence, the wavelength of the wave is equal to 70 m.

8 0
3 years ago
A useful tool for recoding observations for later analysis is a. This is for science
Gnoma [55]
What type of recording device? You can have a: camera and voice recorder, these are the only type of recording device I can think of

There may be more, but these are the ones I think you are allowed... hope this helps you ☁︎☀︎☁︎
5 0
3 years ago
Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill
laila [671]

Answer:

a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

        θ = sin⁻¹ ( λ  / 3d)

the fourth maximum m = 4

       θ = sin⁻¹ ( λ  / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

       I = I₀ cos² (Ф) (sin x / x)²

       Ф = π d sin θ /λ

       x = pi a sin θ /λ

where a is the width of the slits

with the values ​​of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference       d sin θ = m λ

first diffraction minimum    a sin θ = λ

we divide the two expressions

                       d / a = m

In our case

                   3a / a = m

                    m = 3

order three is no longer visible

7 0
2 years ago
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