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3 years ago

Non-metals can form cations when combined with metals and anions when combined with other non-metals.

2 answers:

5 0

I believe it's false because a lot<span> of products nowadays own and use these metals in their product.</span><span>
</span><span>
</span><span>Hope This Helps!!! :)</span>

exis [7]3 years ago

4 0

Well i think it is false caus ebunch of products these day have metals incorparded in the product

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what is the major cause of the earth’s magnetic field? the earth’s liquid outer core the earth’s liquid outer core the earth’s m

GaryK [48]

The earth's liquid outer core is the major cause of the earth’s magnetic field.

<h3>What is magnetic field?</h3>

The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, a vector field. A force acting on a charge while it travels through a magnetic field is perpendicular to both the charge's motion and the magnetic field. The magnetic field of a permanent magnet attracts or repels other magnets as well as ferromagnetic elements like iron. A magnetic field that varies with location will also exert a force on a variety of non-magnetic materials by changing the velocity of those particles' outer electrons. Electric currents, like those utilised in electromagnets, and electric fields that change over time produce magnetic fields that surround magnetised things.

To learn more about magnetic field,visit:

brainly.com/question/11514007

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4 0

2 years ago

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

AlladinOne [14]

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0

3 years ago

The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point

makvit [3.9K]

To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.

5 0

3 years ago

How should scalar symbols be written? A. in bold type B. in italic type C. with an arrow on top D. in bold type or with an arrow

Gemiola [76]

D is the correct answer..........

8 0

3 years ago

A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,

beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

$\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2$

∴ Average intensity is 0.477 W/m²

b) Rms value

$\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C$

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

$E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C$

∴ Peak value of the electric field is 18.96 N/C

8 0

3 years ago