What does it mean to say that two systems are in thermal equilibrium

With a force of 5 newtons, Amanda pushes the stack of books to the right. At the same time, Jeremy, her little brother, pushes t

Vaselesa [24]

Its letter C. 5N to the left. Since Jeremy's force in Newtons are higher than Amanda's (in newtons), and since Jeremy's force directs to the left, then the direction of the force will be to the LEFT. Then subtract the higher one to the lower one so that would be: 10N-5N=5N. So it is C. 5N to the left.

5 0

3 years ago

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A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto

igor_vitrenko [27]

a) The angular velocity of the electron is $4.12\cdot 10^{16} rad/s$

b) The number of revolutions per second is $6.54\cdot 10^{15}$ rev/s

c) The centripetal acceleration of the electron is $8.98\cdot 10^{22} m/s^2$

Explanation:

a)

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

$\omega = \frac{v}{r}$

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the angular velocity is

$\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s$

b)

The period of revolution of the electron is given by

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

where

$\omega = 4.12\cdot 10^{16}rad/s$ is the angular velocity

Substituting,

$T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s$

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

$f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}$ rev/s