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2 years ago

6

A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally

rotating at 55 rpm. What was the initial rotation rate in radians per second?

1 answer:

nadezda [96]2 years ago

3 0

Answer: 5.76 rads/s

Explanation:

The initial rotation is 55 rpm

1 rev = 2π radians

55 revs = 55 × 2π/1 = 345.58 radians/min

345.58 rads/min = 345.58rads/60s = 5.76 rads/s

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Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha

devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_{max}=\frac{hc}{\lambda}-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{495*10^{-9}m}-0.5eV\\W=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0

2 years ago

Un niño de 25 kg corre por un jardín con una velocidad de 2.5 m/s de forma que su trayectoria es tangente al borde de un carruse

Schach [20]

Answer:

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

Explanation:

Asumamos que tanto el niño como el carrusel no tienen carga externa aplicada sobre aquellos, de modo que se puede aplicar el Principio de Conservación de la Cantidad de Movimiento Angular:

$m\cdot v \cdot R = (m\cdot R^{2}+I)\cdot \omega$ (1)

Donde:

$m$ - Masa del niño, medida en kilogramos.

$v$ - Velocidad lineal inicial del niño, medida en metros por segundo.

$R$ - Radio máximo del carrusel, medida en metros.

$I$ - Momento de inercia del carrusel, medida en kilogramo-metros cuadrados.

$\omega$ - Velocidad angular final del sistema niño-carrusel, medida en radianes por segundo.

Si sabemos que $m = 25\,kg$ , $v = 2.5\,\frac{m}{s}$ , $R = 2\,m$ y $I = 500\,kg\cdot m^{2}$ , tenemos que la velocidad angular final es:

$\omega = \frac{m\cdot v\cdot R}{m\cdot R^{2}+I}$

$\omega = \frac{(25\,kg)\cdot \left(2.5\,\frac{m}{s} \right)\cdot (2\,m)}{(25\,kg)\cdot (2\,m)^{2}+500\,kg\cdot m^{2}}$

$\omega = 0.208\,\frac{rad}{s}$

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

4 0

2 years ago

A cow’s mass is 410 kg and a car’s mass is 565 kg. What is the difference between their weights?

solmaris [256]

B. 1520 is the difference between their weights.

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Which field in an 802.11a plcp frame are used to initialize part of the transmitter and receiver circuits?

defon

What Class Is this for?

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Vikki [24]

1 isla thanks for your help and I hope you are feeling 44356

7 0

1 year ago