A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

Substituting the values into the equation

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

Insert the values into the formula

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

Answer:

a)n= 3.125 x electrones.

b)J= 1.515 x A/m²

c) =1.114 x m/s

d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

diámetro 'd'= 2.05 x m

radio 'r' = d/2 => 1.025 x m

número de electrones 'n'= 8.5 x

a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

Como sabemos que: Q= ne

donde e es la carga del electrón, es decir, 1.6 x C

n= Q/e => 5/ 1.6 x

n= 3.125 x electrones.

b) La densidad de corriente 'J' en el cable se calcula como

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x )²)

J= 1.515 x A/m²

c) La velocidad típica '' de un electrón se expresa como:

=

    =1.515 x / 8.5 x x |-1.6 x |

=1.114 x m/s

d) De acuerdo con estas ecuaciones,

J= I/A

= =

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

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Thanks for asking your question here. I hope this response provides clarity. Feel free to ask additional questions. The moment resulting from the two forces about point O is 376 lb-ft counterclockwise.

Response:

length = 2L, mass = M/2, and maximum angular displacement = 1 degree

Clarification:

We examine only small amplitude oscillations (as in this scenario), which keeps the angle θ sufficiently small. In such situations, it's important to note that the pendulum's motion can be described by the equation:

The resulting solution is:

Here, represents the angular frequency of the oscillations, enabling us to find the period:

As a result, the period of a pendulum is determined solely by its length and is independent of both its mass and angle, provided the angle remains small. Therefore, the choice with the longest length gives the longest period.