Trevor places a mass on a spring. When he releases it, the mass and spring move in simple harmonic motion.

Physics

2 answers:

777dan777 [17]3 years ago

8 0

The correct answer is D He could pull the mass down farther.

Amplitude, is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. So to increase the amplitude the spring should be pulled down further which can increase the amplitude.

Elis [28]3 years ago

8 0

The answer is D, He could pull the mass down further.

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Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

marshall27 [118]

Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.

Entonces tenemos que:

$Vf ^{2} -Vo ^{2} =2 x g x h$

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

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Que tengas un buen día!

6 0

3 years ago

Calculate the power of convex lens of focal length 20cm

maxonik [38]

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3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

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Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$