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asambeis [7]
3 years ago
6

An engine has an energy input of 125j and 35 j of that energy is transformed into useful energy

Physics
1 answer:
Tatiana [17]3 years ago
5 0
The engine's efficiency is (35J)/125J) = 28% .

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Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m
Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

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distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

so

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

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