Which force requires contact?

14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change

klio [65]

Lets se

And

$\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}$

$\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}$

So

$\\ \rm\Rrightarrow k\propto m$

If spring constant is doubled mass must be doubled

8 0

2 years ago

Se o Universo tem uma idade definida, de aproximadamente 13,7 bilhões de anos, o que existiu antes do big-bang?

UkoKoshka [18]

Ninguém sabe exatamente

4 0

3 years ago

A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the

Mandarinka [93]

Answer:

a) 0.048A

b) 0.18µC

c) 1.85A

Explanation:

The discharged current of the capacitor as a function of time is given by:

$i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\$

$\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s$

a)

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A$

b)

$q=q_o*e^{-\frac{t}{\tau}}$

$q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC$

c) the maximum current occurs when t=0

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A$

6 0

3 years ago

Verify that the linear speed of an ultracentrifuge is about 0.50 km's, and Earth in its orbit is about 30 km/s by calculating:

FrozenT [24]

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle ( $v$ ), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

$v = R\cdot \omega$ (1)

Where:

$R$ - Radius, measured in kilometers.

$\omega$ - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that $\omega \approx 5235.988\,\frac{rad}{s}$ and $R = 1\times 10^{-4}\,km$ , then the linear velocity is:

$v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)$

$v = 0.524\,\frac{km}{s}$

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

$\omega = \frac{2\pi}{T}$ (2)

Where $T$ is the period, measured in seconds.

If we know that $T = 31536000\,s$ , then the angular speed of the Earth is:

$\omega = \frac{2\pi}{31536000\,s}$

$\omega = 1.992\times 10^{-7}\,\frac{rad}{s}$

Now, we determine the linear speed:

$v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)$

$v = 29.88\,\frac{km}{s}$

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

6 0

2 years ago

Different between pressure and force

9966 [12]

Force is mass into acceleration

and pressure is force applied per unit area.

3 0

3 years ago