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If an X-ray tube is operating at a current of 30.0 mA:a) How many electrons are striking the target per second? b) If the potential difference between the anode and cathode of this tube is 100.0 kV, how much power is expended

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hopemichael95

Answer:

Explanation:

A. Q= It

30E-3A=q

Ne = q

30E-3/1.6*10-19= N

N= 1.8*10^16 electrons

B. Power= I x v

= 30*10^-3A x 100*10^3v

= 3000watts

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