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4 years ago

5

____ developed the first model of the atom that showed the structure of the inside of an atom. A. Dalton B. Bohr C. Rutherford D

. Thomson

1 answer:

Elan Coil [88]4 years ago

6 0

Answer:

D. Thomson

Explanation:

The first model of the atom was proposed by J. J. Thomson.

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List two pieces of evidence<br> related to this picture

tatyana61 [14]

Answer:

there was a crash you can tell cause of the 2 cars the dog is curious on what happened, you can tell because the dog is looking at the cash

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3 years ago

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Which example describes voluntary muscle movement?

Leni [432]

A voluntary muscle movement would be like purposely flexing you arm to show off muscles or to use your muscles to do a push up. Basically when you chose to do something

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3 years ago

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A horizontal 791-N merry-go-round is a solid disk of radius 1.56 m and is started from rest by a constant horizontal force of 49

ICE Princess25 [194]

Answer:

$K.E=273.5J$

Explanation:

Given data

$F_{g}=791N\\F_{h}=49N\\r=1.56m\\t=3.03s$

To find

Kinetic Energy

Solution

The moment of inertia is given as:

$I=(\frac{1}{2} )MR^2\\I=\frac{1}{2}(F_{g}/g)R^2\\ I=\frac{1}{2}(\frac{791N}{9.8m/s^2} )(1.56m)^2\\ I=98.21kg.m^2$

The angular acceleration is given as:

$\alpha =\frac{T}{I}\\\alpha =\frac{F_{y}R}{I}\\ \alpha =\frac{(49N)(1.56m)}{98.2kg.m^2}\\\alpha =0.778rad/s^2$

Now the angular velocity is given by:

$w=\alpha t\\w=(0.778rad/s^2)(3.03s)\\w=2.36rad/s$

So the kinetic energy given as:

$K.E=(\frac{1}{2} )Iw^2\\K.E=\frac{1}{2}(98.21kg.m^2)(2.36rad/s)^2\\ K.E=273.5J$

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3 years ago

(a) calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the radius of the earth is

kakasveta [241]

(a) The value of the gravitational acceleration is given by:

$g=\frac{GM}{r^2}$

where

$G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

M is the Earth's mass

$r=6371 km=6.371 \cdot 10^6 m$ is the Earth's radius at the pole

If we use the value for g given by the problem, $g=9.830 m/s^2$ , and we rearrange the equation above, we find the value of the Earth's mass:

$M=\frac{gr^2}{G}=\frac{(9.830 m/s^2)(6.371 \cdot 10^6 m)^2}{6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}}=5.982 \cdot 10^{24} kg$

(b) The value we found for the Earth's mass is $5.982 \cdot 10^{24} kg$ , and we see that this value is slightly larger than the accepted value of $5.979 \cdot 10^{24} kg$ .

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3 years ago

What happens to an atom if the electrons in the outer shell are altered?

maxonik [38]

Answer:

NUCLEAR BOMB

Explanation:

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3 years ago

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