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3 years ago

13

What was one main point of Dalton's atomic theory

2 answers:

katovenus [111]3 years ago

5 0

Indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

goblinko [34]3 years ago

4 0

All matter is composed of atoms, indestructible building blocks.

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If i try to fail and succeed which one did I do

Novay_Z [31]

You've failed because you failing becomes a statement rather than it becoming fact or what actually happened.

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3 years ago

9. A wave on Beaver Dam Lake passes by two docks that are 40.0 m apart.

Kobotan [32]

Answers:

a) 10 m

b) time=1.6 s, frquency=0.625 Hz

c) 6.25 m/s

Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength $\lambda$ is the distance between to crests; this means we have $4\lambda$ in $40 m$ :

$40 m=4\lambda$

Clearing $\lambda$ :

$\lambda=\frac{40 m}{4}$

$\lambda=10 m$

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- $T$

Then:

$T=\frac{(1 wave)(16 s)}{10 waves}$

$T=1.6 s$ This is the period of the wave

On the other hand, the frequency $f$ of the wave has an inverse relation with its period $T$ :

$f=\frac{1}{T}$

$f=\frac{1}{1.6 s}$

$f=0.625 Hz$ This is the frequency of the wave

c) The speed $v$ of a wave is given by the following equation:

$v=\frac{\lambda}{T}$

$v=\frac{10 m}{1.6 s}$

Finally:

$v=6.25 m/s$

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3 years ago

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mihalych1998 [28]

Explanation:

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2 years ago

Read 2 more answers

Writing a 4 to 5 sentence paragraph, identify YOUR stance on whether more Nuclear Reactors should be built here in Michigan. (1s

prisoha [69]

Thank You for giving me the points ^_^

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2 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago