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3 years ago

GOOD POINTS HELP PLEASE IN SCIENCE

Physics

1 answer:

jeka943 years ago

6 0

The answer is c because a metallic bond Is 1. formed of the attraction between positively charged metal nuclei
2. and surrounding sea electrons

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what is the mass of vertical column of air that exerts pressure of 1 atm over a10m by 10m square surface

Oksana_A [137]

Answer:

$m=1.01\times 10^6\ kg$

Explanation:

Given that,

Pressure, P = 1 atm = 101325 Pa

Area of the square surface, A = 10² = 100 m²

We need to find the mass of vertical column of air. We know that, pressure is equal to the force acting per unit area. So,

$P=\dfrac{mg}{A}\\\\m=\dfrac{PA}{g}\\\\m=\dfrac{101325\times 10^2}{10}\\\\m=1.01\times 10^6\ kg$

So, the required mass of the vertical column of air is $1.01\times 10^6\ kg$ .

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3 years ago

ILL GIVE BRAINLIEST!! URGENT PLEASE HELP

anygoal [31]

I believe the answer is the second one.

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3 years ago

Any two waves that meet as they travel through the same medium will interact. If the two waves are in phase their crests will co

MrRissso [65]

Answer:

constructive

alpitude

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3 years ago

Lena helps a child who is choking on his food. Which reliable method prepared her for helping with this injury?

Ivanshal [37]

knowledge of first aid ... eg St John Ambulance, Red Cross etc. I think that everyone in a school should be taught First Aid.

7 0

3 years ago

Read 2 more answers

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago