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3 years ago

What was one main point of Dalton's atomic theory

2 answers:

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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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

PLEASE ANSWER! WILL MARK BRAINLIEST!

Angelina_Jolie [31]

The first one would be thermal energy

8 0

3 years ago

which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload

Anuta_ua [19.1K]

Overload <<<<<<<<<<<<<<<<<<<<<<<<<<<<<,,,,,,,,

5 0

3 years ago

Which vector is the sum of vectors a and ? b а O A. B. C.

seropon [69]

Answer:

i would say D.

Explanation:

5 0

3 years ago

A ball of mass 0.50 kg is fired with velocity 160 m/s into the barrel of a spring gun of mass 1.8 kg initially at rest on a fric

shepuryov [24]

Answer:

All fraction of kinectic energy is lost to barrel of a spring gun of mass 1.8 kg

Explanation:

A ball of mass 0.50 kg is fired with velocity 160 m/s ...

The kinetic energy is given by 1/2mv²

Kinectic energy of the ball = 1/2 *0.5*160²

Kinectic energy = 1/4 *25600

Kinectic energy = 6400 joules.

If no energy is lost to fiction, and the ball sticks to a barrel of a spring gun of mass 1.8 kg with initial velocity zero, all kinetic energy is lost to the barrel of a spring gun of mass 1.8 kg.

5 0

3 years ago