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Flura [38]
3 years ago
13

What they did to expand the atomic theory ?

Physics
1 answer:
Levart [38]3 years ago
7 0

Answer:Dalton just expanded on the Greek idea of the atom.

Explanation:

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A wave with a period of 0.008 second has a frequency of?
tatyana61 [14]
The formula for frequency is f = 1/T where f is frequency and T is period in seconds. 
You have you period which is 0.008s and that is all you will need to solve or frequency in a wave:
f = 1/2
f = 1/0.008s
f = 125Hz
6 0
3 years ago
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What is the formula to find how far a wheel travels in one rotation?
alekssr [168]

it is also known as formula for circumfrence which is 2 times pi times radius. if radius was 5 then the circumfrence would be 10pi.

4 0
3 years ago
What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan
kramer
I think the question should be the below:

<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:

 <span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>

<span>x (max) = a(max) /ω² </span>

<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
5 0
3 years ago
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
When compared to
ElenaW [278]
I think it is B, because the sun’s size is pretty average
3 0
2 years ago
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