Hi there! :)
Reference the diagram below for clarification.
1.
We must begin by knowing the following rules for resistors in series and parallel.
In series:
In parallel:
We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.
Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
2.
We can use Ohm's law to solve for the current in the circuit.
3.
For resistors in series, both resistors receive the SAME current.
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.
4.
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.
Answer:
Sarah is right
Explanation:
This is an exercise that differentiates between scalars and vectors.
A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.
In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero
consequently Sarah is right
A. True. You can use displacement to determine the volume of solids and liquids.
Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
<em>Q = mcΔT</em>
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J