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bogdanovich [222]
3 years ago
15

What new characteristics did john dalton add to the model of the atom

Physics
1 answer:
kicyunya [14]3 years ago
7 0

Answer:

All atoms of an element are identical. The atoms of different elements vary in size and mass. Compounds are produced through different whole-number combinations of atoms. A chemical reaction results in the rearrangement of atoms in the reactant and product compounds.

HOPE THIS HELPED!!!!!!!!!!!!!!!!!!! XDDDDDDDDDDD

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A billiard ball collides with a second identical ball in an elastic head-on collision. What is the kinetic energy of the system
murzikaleks [220]

Answer:

Explanation:

From the question, we were made to understand that the collision between the two billiard balls was an elastic collision. Hence, an elastic collision is one in which the kinetic energy is conserved. Meaning the kinetic energy before the collision is still retained after the collision.

Kinetic energy before collision = kinetic energy after collision

1/2mv^2 = 1/2mv^2

There was no gain nor loss in energy

5 0
3 years ago
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An apple suddenly drops a distance of 3.2 m from a tree. If the acceleration due to gravity is 9.8 m/s2, how long does it take t
mart [117]

Answer:

0.33 seconds

Explanation:

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3 years ago
How are the electric field lines around a positive charge affected when a second positive charge is near it? The field lines com
timofeeve [1]
The answer would be D hope it helps and sorry if it is wrong.  :)
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3 years ago
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PLEASE HELP!!! 25 pts!!
mina [271]

Answer:

The slope of the graph is what you need. That tells you the speed not the velocity. In order to find the velocity you would also need to know the direction of the motion.

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3 years ago
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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
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