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Sagot :
To determine how much power is produced by the flashlight, we can use the formula for electrical power:
[tex]\[ \text{Power} (P) = \text{Voltage} (V) \times \text{Current} (I) \][/tex]
Given the values:
- Voltage ([tex]\( V \)[/tex]) = 12 volts
- Current ([tex]\( I \)[/tex]) = [tex]\( 6.5 \times 10^{-2} \)[/tex] amps
We substitute these values into the formula:
[tex]\[ P = 12 \, \text{volts} \times 6.5 \times 10^{-2} \, \text{amps} \][/tex]
Calculating this gives us:
[tex]\[ P = 12 \times 0.065 \][/tex]
Therefore:
[tex]\[ P = 0.78 \, \text{watts} \][/tex]
So, the correct answer is:
A. 0.78 watts
[tex]\[ \text{Power} (P) = \text{Voltage} (V) \times \text{Current} (I) \][/tex]
Given the values:
- Voltage ([tex]\( V \)[/tex]) = 12 volts
- Current ([tex]\( I \)[/tex]) = [tex]\( 6.5 \times 10^{-2} \)[/tex] amps
We substitute these values into the formula:
[tex]\[ P = 12 \, \text{volts} \times 6.5 \times 10^{-2} \, \text{amps} \][/tex]
Calculating this gives us:
[tex]\[ P = 12 \times 0.065 \][/tex]
Therefore:
[tex]\[ P = 0.78 \, \text{watts} \][/tex]
So, the correct answer is:
A. 0.78 watts
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