Answer:
B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Explanation:
Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.
Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.
So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.
Also,option D is ruled out as ammeter is connected in series.
Answer:
I = 4.75 A
Explanation:
To find the current in the wire you use the following relation:
(1)
E: electric field E(t)=0.0004t2−0.0001t+0.0004
ρ: resistivity of the material = 2.75×10−8 ohm-meters
J: current density
The current density is also given by:
(2)
I: current
A: cross area of the wire = π(d/2)^2
d: diameter of the wire = 0.205 cm = 0.00205 m
You replace the equation (2) into the equation (1), and you solve for the current I:
Next, you replace for all variables:
hence, the current in the wire is 4.75A
Answer:
a. get warmer.
Explanation:
When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.
It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.
Answer:
Je ne Sachez que Qu’est-ce que le
Answer:
Scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis.
Explanation:
