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weeeeeb [17]
2 years ago
11

How many volts would it take to push 1 amp through a resistance of 1 ohm?

Physics
2 answers:
ELEN [110]2 years ago
5 0
V=I x R so V= 1 x 1 =1V
Diano4ka-milaya [45]2 years ago
4 0

Answer:

1\; \rm V.

Explanation:

Ohm's Law give the following relationship between the potential V across a conductor, the electrical resistance R of that conductor, and the current I through that conductor:

V = I \, R.

It is given that the current through this conductor is I = 1\; {\rm A}, and that the electrical resistance of this conductor is R = 1\; \rm \Omega.

Note, that the unit of electrical resistance Ohm is a composite unit; 1\; {\rm \Omega} = 1\; {\rm V \cdot A^{-1}} (volts per ampere.)

Substitute the value of I and R into the equation V = I \, R from Ohm's Law to find the potential across this conductor:

\begin{aligned}V &= I \, R \\ &= 1\; {\rm A} \times 1\; {\rm \Omega} \\ &= 1\; {\rm A} \times 1\; {\rm V \cdot A^{-1}} \\ &= 1\; {\rm V}\end{aligned}.

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  • Pressure = 5000 Pa
  • Contact Area = 0.04 m^2
  • Acceleration due to gravity = 9.8 m/s^2
  • Let the force be F.
  • We know, Force = Pressure × Contact Area
  • Therefore, Force = 5000 Pa × 0.04 m/s^2
  • or, Force = 200 N
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6 0
2 years ago
The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the
vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r²             where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

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