The balloon could not rise. The air inside would not be an lighters than the air
outside.
Answer:
I think
when the problem has numerous possible options.
If wrong correct me pls
<em><u>Have</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>great</u></em><em><u> </u></em><em><u>day</u></em><em><u> </u></em>
<em><u>I</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>helps</u></em><em><u> </u></em>
<em><u></u></em>
Answer:
0.3 amps
Explanation:
amps can be found using the equation
current=volts/resistance
so current=3V/10Ohms=0.3 amps
Answer:
someone who has sufficient training and experience or knowledge and other qualities that allow them to assist you properly.
Explanation:
Answer:
A. 7199.55 volts
B. 120A
Explanation:
In this question we have the
line voltage = VLL = 12470volts
Phase current = Iph = 120 amps
A.)
We are to calculate the line-to-neutral/phase voltage here
VLL = √3VL-N
VL-N = VLL/√3
VL-N = 12470/√3
This gives a line to neutral phase/voltage of 7199.55 volts.
B.
We are to calculate the line current here:
In this connection, the line current and the phase current are equal
ILL = Iph = 120A