What is the function of the muscles around the lens in the human eye

When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Star

Gemiola [76]

Answer:

α = 200*π rad/sec²

Explanation:

If the disk drive revolves once every 0.05 sec, this means that the angular velocity can be expressed as follows:

$\omega = \frac{2*\pi}{0.05s} = 40*\pi rad/sec$

(just applying the definition of angular velocity)
Now, it the disk started from rest, and took two revolutions to reach to full speed, assuming that the angular acceleration is constant, we can find the angular acceleration applying this kinematic equation:

$\omega_{f} ^{2} -\omega_{0} ^{2} = 2*\alpha * \Delta\theta$

where Δθ, is the angle rotated while it went from rest to full speed, i.e., 2 revolutions, or 2*π rads.
Replacing by the value that we found for ωf (as ω₀ = 0), we can solve for α :

$\alpha =\frac{\omega_{f}^{2} }{2*\Delta\theta} = \frac{(40*\pi)^{2} }{2*2*2*\pi } =\\ \\ \alpha = 200*\pi rad/sec2$

3 0

3 years ago

A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms

Alchen [17]

R(parallel) = product/ sum

50×30/50+30

1500/80

18,75 ohms

4 0

2 years ago

Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su

dybincka [34]

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua ( $v$ ), en metros por segundo:

$v = \sqrt{\frac{K}{\rho} }$ (1)

Donde:

$K$ - Módulo de compresibilidad, en newtons por metro cuadrado.
$\rho$ - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que $\rho = 1\times 10^{3}\,\frac{kg}{m^{3}}$ y $K = 2,16\times 10^{9}\,\frac{N}{m^{2}}$ , entonces la velocidad de las ondas sonoras es:

$v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }$

$v\approx 1469,694\,\frac{m}{s}$

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda ( $\lambda$ ), en metros, mediante la siguiente fórmula:

$\lambda = \frac{v}{f}$ (2)

Donde $f$ es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que $v\approx 1469,694\,\frac{m}{s}$ y $f = 1000\,hz$ , entonces la longitud de onda de las ondas sonoras es:

$\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}$

$\lambda = 1,470\,m$

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

6 0

2 years ago

Give at least one fact about subatomic particles

victus00 [196]

Answer:

- Particles smaller than atoms are called subatomic particles .

- There are three famous subatomic particles, proton, neutron and electron .

- The study of sub atomic particles are called particle physics

- These particles can be divided as Brayons and Leptons

- These particles are often held together by one of the four fundamental particles ( Weak force, strong force, electromagnetic force, gravitational force).

6 0

2 years ago

BRAINLIESTTT!! HELP ME!!!

likoan [24]

The efficiency of the scissor is 200%.

<u>Explanation:</u>

Efficiency is defined as the ratio of output of any instrument or device or machine to the input supplied to it. So the greater the output the greater will be the efficiency of the device.

As here the work done by us on the system is said to be 10 J so this will be equal to the input work done on the system. And the work done by the system i.e., the scissor is 20 J, so this will be the output work.

So, the efficiency is the ratio of output to input as shown below.

Efficiency = $(\frac {20} {10} ) \times 100$ = 200

So, the efficiency of the scissor is 200%.

6 0

3 years ago