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4 years ago

Is it possible for an atom to have an electric charge?

Physics

1 answer:

jeka57 [31]4 years ago

7 0

False an atom can not have an electric charge.

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¿Por qué el origen de ciertos metales se relaciona con la dinámica terrestre?

Lemur [1.5K]

Yo no hablo espanol

5 0

2 years ago

When the mass of an object decreases, the force of gravity

Viktor [21]

Hello There!

From what i know, gravitational force increases if the mass is increased.
If the mass is being decreased, then i assume it will be B. Decreases.

Hope This Helps You!
Good Luck :)

- Hannah ❤

8 0

4 years ago

Calculate the pH of hydronium ions in a 0.000 01 M solution of the strong acid HI.

enyata [817]

<h3>Answer:</h3>

pH = 5

<h3>Explanation:</h3>

<u>We are given;</u>

Concentration of a solution HI as 0.00001 M

We are required to calculate the pH of Hydronium ions.

When the acid dissociates in water to;

HI + H₂O → H₃O⁺(aq) + I⁻(aq)

Therefore;

The concentration of H₃O⁺ ions is 0.00001 M\

We need to know that pH = -log[H₃O⁺]

Therefore;

pH =-log 0.00001 M

= 5

Thus, the pH of the hydronium ions is 5

7 0

3 years ago

Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load

rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

$\sigma=E*\epsilon$ (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

$\delta L=L*\epsilon$ (2)

Here L is the member extension and δL the change total longitudinal elongation.

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

$\sigma=P/A$

$\sigma=22.44N / 1290 mm^2$

$\sigma=0.0174 N/mm^2$

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

$=\sigma=E*\epsilon$

$\epsilon=\sigma/E$

$\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2}$

$\epsilon=8.53*10^-{5}$

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

$\delta L=3048 mm * 8.53*10^{-5}$

$\delta L= 0.25 mm$

4 0

3 years ago

A duck is floating on a lake with 25% of its volume beneath the water. what is the average density of the duck?

Mama L [17]

Answer: 250kg/m3

The duck condition would be 25% on water and 75% on air. That means the duck density is lower than the water and the duck is pushed up by the force. The force difference from 25% of duck volume can sustain 100% of the duck weight.
That mean : density of the duck/ density of water = 25%/100%= 0.25 x water density.
If the density of water is 1000kg/m3 then duck density : 0.25x 1000kg/m3= 250kg/m3

4 0

4 years ago