If a bar of copper is brought near a magnet, the copper bar will be unaffected by the magnet.
The sample appears to have gone through 3 half-lives
1st half life: 1000 to 500 g
2nd half life: 500 to 250 g
3rd half life: 250 to 125 g
The duration of a half-life, therefore, can be inferred to be 66 ÷ (3) = 22 days.
After a 4th half life, there will be 125÷2= 62.5 g.
At this point, an additional 22 days will have passed, for a total of 88 days.
Answer is C.
Answer:
The electric current through the resistor is: 7.66667 milliamps
Explanation:
Applying Ohm's law we have or solving to so now, we can replace the values given before in the equation, we need to know that the voltage across the resistor is which determine the current in the component, in this case, the resistor. After clear this topic, we get so 1 milliamps is 0,001 amps we get 7,66667 milliamps.
The correct answer is 100 N.
In fact, the problem already says that the force exerted by magnet A on magnet B is exactly 100 N. The information about the magnetic field intensity of the two magnets is redundant, because we are interested only in the magnitude of the force.
Answer:
0
-723.163841808 Wb
-350.282485876 Wb
Explanation:
= Permittivity of free space =
when r = 0.625 m the charges lie outside the sphere so q = 0
From Gauss law
The electric flux is 0
when r = 1.4
will be inside the sphere
The electric flux is -723.163841808 Wb
when r = 2.9 both charges lie inside the sphere
The electric flux is -350.282485876 Wb