When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th
2 answers:
The system undergoes 20 full vibrations within a 4-second interval, resulting in a frequency of
The angular frequency can be determined as
In simple harmonic motion, the angular frequency is also described by
where k represents the spring constant, and m=25 g is equivalent to 0.025 kg, the mass connected to the spring. Utilizing the previously calculated angular frequency, we can establish the value of k:
The angular frequency can be determined as
In simple harmonic motion, the angular frequency is also described by
where k represents the spring constant, and m=25 g is equivalent to 0.025 kg, the mass connected to the spring. Utilizing the previously calculated angular frequency, we can establish the value of k:
Response: The spring constant is 25 N/m.
Details:
The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).
The total oscillations are 20 in 4 seconds.
Oscillations per second =
Spring's frequency of vibration is =
The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.
The spring constant is 25 N/m.
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Answer:
La magnitud del momento total es de 21.2 kg m/s y su dirección es de 39.5° respecto al eje x.
Explanation:
¡Hola!
El momento total se calcula como la suma de los momentos de las piezas.
El momento de cada pieza se calcula de la siguiente manera:
p = m · v
Donde:
p = momento.
m = masa.
v = velocidad.
El momento es un vector. La pieza de 200 g se mueve a lo largo del eje x, por lo que su momento será:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
La pieza de 300 g se mueve a lo largo del eje y. Su vector momento será:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
El momento total es la suma de cada momento:
Momento total = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Momento total = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Momento total = (16.4 kg m/s, 13.5 kg m/s)
La magnitud del momento total se calcula de la siguiente manera:
La dirección del vector de momento se calcula utilizando trigonometría:
cos θ = px/p
Donde px es el componente horizontal del momento total y p es la magnitud del momento total.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° si no redondeamos la magnitud del momento total)
<pFinalmente, la magnitud del momento total es 21.2 kg m/s y su dirección es 39.5° respecto al eje x.
Answer:
Amplitude, A = 0.049 meters
Explanation:
Given that
A harmonic wave propagates in the positive x direction at 6 m/s along a tight string. A fixed point along this string oscillates over time according to the equation:
.......(1)
The general wave equation is expressed as:
.......(2)
A denotes the wave's amplitude
When we compare equation (1) with (2), we find:
A = 0.049 meters
Thus, the amplitude of the wave is 0.049 meters.
Response:
The primary consequence is an increase in induced charge at the nearest points. However, the overall net charge remains zero, meaning it does not influence the flow.
We can utilize Gauss's law to solve this problem
Ф = ∫ e. dA = / ε₀
The flow of the field is directly correlated to the charge within it. Consequently, placing a Gaussian surface beyond the non-conductive spherical shell means the flow will be zero since the sphere’s charge equals the charge induced in the shell, resulting in a net charge of zero. This evaluation shows that the shell effectively obstructs the electric field.
According to Gauss's law, if the sphere is offset, the only effect it generates is an increment in induced charge at the nearest points. Nevertheless, the net charge remains zero, so it does not impact the flow; irrespective of the sphere's position, the total induced charge is consistently equal to the charge on the sphere.