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3 years ago

Which best describes the current atomic model?

Physics

2 answers:

andreev551 [17]3 years ago

7 0

Answer:

Explanation:

It is not an exact representation of the atom, but is close enough to be very useful.

MissTica3 years ago

5 0

It is not an exact representation of the atom, so it is not very useful.

You might be interested in

A 75.0-kg person climbs stairs, gaining 2.50 m in height. Find the work done to accomplish this task.

ololo11 [35]

Answer:

$W=1837.5J$

Explanation:

A force exerts work when there is a displacement of its point of application in the direction of that force. Therefore, the work done by a system is defined as the inner product between the applied force and the displacement:

$W=\vec{F}\cdot \vec{d}\\W=Fcos\theta d$

In this case, we have:

$F=mg\\h=dcos\theta$

So, replacing this:

$W=mgh\\W=75kg*9.8\frac{m}{s^2}*2.5m\\W=1837.5J$

4 0

3 years ago

Suppose a moving car has 2000 J of kinetic energy. If the carʹs speed doubles, how much kinetic energy would it then have?

Bond [772]

Answer:

option A

Explanation:

given,

Kinetic energy of the car = 2000 J

speed of the car is doubled

we know,

$KE_1 = \dfrac{1}{2}mv^2$

$2000= \dfrac{1}{2}mv^2$ ........(1)

now, speed of the car is doubled

v' = 2 v

$KE_2 = \dfrac{1}{2}mv'^2$

$KE_2 = \dfrac{1}{2}m(2v)^2$

from equation (1)

$KE_2 = 4\times \dfrac{1}{2}m(v)^2$

$KE_2 = 4\times 2000$

$KE_2 = 8000\ J$

Hence, the Kinetic energy would be equal to 8000 J.

The correct answer is option A.

8 0

2 years ago

The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is ob

Naddik [55]

(a) $0.2888 kg m^2$

The moment of inertia of a uniform-density disk is given by

$I=\frac{1}{2}MR^2$

where

M is the mass of the disk

R is its radius

In this problem,

M = 16 kg is the mass of the disk

R = 0.19 m is the radius

Substituting into the equation, we find

$I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2$

(b) 142.5 J

The rotational kinetic energy of the disk is given by

$K=\frac{1}{2}I\omega^2$

where

I is the moment of inertia

$\omega$ is the angular velocity

We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s$

And so, the rotational kinetic energy is

$K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J$

(c) $9.07 kg m^2 /s$

The rotational angular momentum of the disk is given by

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular velocity

Substituting the values found in the previous parts of the problem, we find

$L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s$

8 0

3 years ago

What is evidence (and reasoning) on the Earth beginning as a small dense hot point (and beginning "13.7 billion years ago")

Taya2010 [7]

None whatsoever.
You're confusing the formation of the Earth with the so-called "Big Bang".
The sun and planets formed billions of years AFTER the Big Bang.

4 0

2 years ago

Winding a chain A 30-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the syste

Tcecarenko [31]

Answer:

Explanation:

Assuming that velocities remain negligible,

The work will be equal to the change in potential energy of the center of mass

mass is 30 m(5 kg/m) = 150 kg

W = 150(9.8)(30/2) = 22,050 J ≈ 22 kJ

5 0

2 years ago