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Blababa [14]
3 years ago
9

Why did Dalton think it was important to use his system of symbols for the chemical elements.

Physics
1 answer:
STALIN [3.7K]3 years ago
4 0
It was essential to remind people that all the matter around us has one base unit: the atom. Dalton thought it was important to use his system of symbols for the representation of chemical elements because it was helpful to remember exactly which atoms lay at the basic structure of the matter. It was easy for people to use symbols for elements in chemical formulas. He pictured the atoms as small balls and this is the same way that he depicted models of elements and compounds. 
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How resistance, current, and voltage behave in a series circuit.?
mylen [45]

b.the covalent circuit is held throw magnetic waves


4 0
4 years ago
(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar
likoan [24]

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

7 0
3 years ago
Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5
AlladinOne [14]

Answer:

a)432000\frac{N}{C}\hat{i}

b)-6.92\times10^{-14}N\hat{i}

Explanation:

a)

The magnitude of the electric field generated by a charged particle at a distance r is:

E= k\frac{|Q|}{r^{2}}

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is 25.0cm-5.0cm=20.0 cm=0.2m apart the positive charge is:

E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}

E_1= 45000\frac{N}{C}

and the electric field generated by q2:

E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}

E_2=216000\frac{N}{C}

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}

\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}

Vector quantities satisfy superposition principle, this is \overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}, with E the total electric field.

\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}

b) The force is:

\overrightarrow{F}=e*\overrightarrow{E},

with q the charge of an electron

\overrightarrow{F}=(-1.61\times10^{-19})*(432000)\hat{i}=-6.92\times10^{-14}N\hat{i}

8 0
3 years ago
What is the tail of a comet made of
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What is the term for an electrical channel within the circuitry of a computer than can transfer 8 bytes of data at a time?
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4 0
3 years ago
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