The direction of the force experienced by the positive charge is upward.
We can use the right-hand rule to understand the direction of the Lorentz force acting on the charge: let's put the thumb in the same direction of the current in the wire (eastward), while the other fingers "wrap themselves" around the wire. These other fingers give the direction of the Lorentz force in every point of the space around the wire. Since the charge is located north of the wire, in that point the fingers are directed upward, so the positive charge experiences a force directed upward.
(if it was a negative charge, we should have taken the opposite direction)
They point towards a negative charge.
They point away from a positive charge.
They never cross.
Answer:
Explanation:
Givens
Heat of Fusion = 2.05 * 10^5 J / kg watch the units.
Heat to actually melt the copper = 82 10^5 J
Formula
Mass of copper = Heat / Heat of Fusion
Solution
Mass of copper = 82*10^5 J / (2.05 * 10^5 J / kg)
Mass of copper = 40 kg
Notice that the kg is in the denominator of the second fraction. The rules of fractions would tell you the 1/1 / / 1 /kg . You take the right fraction and turn it upside down and multiply. 1 / 1 * kg/1 = 1* kg / 1*1 which is just kg.
Answer 40 kg of copper
Answer:
Approximately 
, assuming that this gas is an ideal gas.
Explanation:
- Let
and 
denote the volume and pressure of this gas before the compression. - Let
and 
denote the volume and pressure of this gas after the compression.
By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:
.
Note that in Boyle's Law, 
is inversely proportional to 
. Therefore, on the two sides of this equation, "final" and "initial" are on different sides of the fraction bar.
For this particular question:
.
.
.- The pressure after compression, , needs to be found.
Rearrange the equation to obtain:
.
Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.
.
