What are five atomic models that have been proposed over time ?

dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0

3 years ago

Which of the following is a type of physical change?

goblinko [34]

Answer is C. All of the others are internal/molecular changes.

8 0

3 years ago

Read 2 more answers

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s

asambeis [7]

Answer:

$I=0.0987kg.m^2$

Explanation:

From the question we are told that:

Mass $M=1.80kg$

Deviation $d=0.250$

Time $t=0.940s$

Generally the equation for moment of inertia is mathematically given by

$I=\frac{T}{2\pi}^2(mgd)$

$I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)$

$I=0.0987kgm^2$

5 0

3 years ago

Describe two experiments to determine the speed of propagation of a transverse wave on a rope. You have the following tools to u

AnnZ [28]

Answer:

#See solution for details.

Explanation:

1.

Tools: $stopwatch, \ meter \ stick, \ mass \ measuring \ scale , \ force \ measuring \ device$ .

$Experiment \ 1$ :Calculate the speed of the wave using the time, $t$ it takes to travel along the rope. Rope's length, $L$ is measured using the meter stick.

-Attach one end of rope to a wall or post, shake from the unfixed end to generate a pulse. Measure the the time it takes for the pulse to reach the wall once it starts traveling using the stopwatch.

-Speed of the pulse can then be obtained as:

$v=\frac{L}{t}$

$Experiment \ 2$ : Apply force of known value to the rope then use the following relation equation to find the speed of a pulse that travels on the rope.

$v=\sqrt{\frac{F}{\mu}}\ ,\mu=\frac{m}{L}$

-Use the measuring stick and measuring scale to determine $L,m$ values of the rope then obtain $\mu$ .

-Use the force measuring constant to determine $F$ . These values can the be substituted in $Experiment \ 1$ to obtain $v.$

4 0

3 years ago

What are the units, if any, of the particle in a box wavefunction. What does this mean?

SIZIF [17.4K]

Answer:

$[\psi]= [Length^{-3/2}]$
This means that the integral of the square modulus over the space is dimensionless.

Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

$\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \, dy \, dx$

must be dimensionless, as represents a probability.

As the differentials has units of length

$[dx]=[dy]=[dz]=[Length]$

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

$[\psi]^2 = [Length^{-3}]$

taking the square root this gives us :

$[\psi] = [Length^{-3/2}]$

5 0

3 years ago