Answer: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below
Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
Answer:
the magnitude of acceleration will be 1.50m/s^2
Explanation:
To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma
if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.
The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).
you will find that Fnet=15.0N
With that, plug in the values you know to calculate the acceleration of the block:
Fnet=ma
(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:
1.50=a (and now make sure to label the units of your answer)
a=1.50m/s^2 (which is the typical unit for acceleration)
-- In a series circuit, the current ( I ) is the same at every point.
-- The power dissipated by any section of the circuit is I² x Resistance.
-- The wire has very low resistance, so I²R is very low dissipated power.
-- The filament in the bulb has most all of the resistance in the circuit,
so it dissipates virtually all the power of the circuit, and certainly much
more than the wires do.