Answer:
Option 3
Explanation:
O Option C is NEGATIVELY CHARGED, meaning it has GAINED ELECTRONS resulting in a GREATER number of ELECTRONS than PROTONS.
Answer:
The electrical force between two balloons is 67.5N.
Explanation:
There are two charged balloons, let's say a and b.
The charge on the balloon a = C
The charge on the balloon b = C
Both balloons are 1 cm apart; it means that the distance<em> r</em> between the balloon a and the balloon b is 0.01 m (since 1 cm = 0.01 m).
We need to find the electrical force between them. By using the Coulomb's law, the magnitude of the electrical force between both the balloon is given as follows:
--- (A)
Where,
k = Coulomb's constant = =
Plug all the values in the equation (A):
Hence, the electrical force between two balloons is 67.5N (three significant figures).
Answer:
Question 1: 189 joules
Question 2: 2.25 watts
Question 3: 405 joules
Question 4: 0.00125 watts
Question 5: 450000 joules
Explanation:
question 1:
Volt=1.5v
Resistance =5 ohms
Time=7 minutes
Time=7 x 60
Time=420 seconds
Current =voltage ➗ resistance
Current =1.5 ➗ 5
Current =0.3 amps
Energy=current x voltage x time
Energy=0.3 x 1.5 x 420
Energy =189 joules
Question 2:
Current =0.5 amps
Voltage =4.5v
Power=current x voltage
Power=0.5 x 4.5
Power=2.25 watts
Question 3:
Current=0.5 amps
Voltage=4.5v
Time =3 minutes
Time =3x60
Time =180 seconds
Energy=current x voltage x time
Energy=0.5 x 4.5 x 180
Energy =405 joules
Question 4:
Resistance=50 ohms
Current =5 milliamps
Current =5/1000
Current =0.005 amps
Power =current x current x resistance
Power=0.005 x 0.005 x 50
Power=0.00125 watts
Question 5:
Power =50 watts
Time=2.5 hour
Time=2.5 x 60 x 60
Time =9000 seconds
Energy=power x time
Energy=50x9000
Energy=450000 joules
It is the third option, buoyant force.
Hi, the answer to this would be the conservation of mass.
Some history on this: The law of the conservation of mass was founded by a french chemist named Antoine Lavoisier, and stated that: matter cannot be created or destroyed in a chemical reaction.
Hope this helps and if you have any more questions you can message me and I'll be glad to help. :)