Answer:
0.36 A.
Explanation:
We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:
Resistor 1 (R₁) = 35 Ω
Resistor 2 (R₂) = 20 Ω
Equivalent Resistance (Rₑq) =?
Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:
Rₑq = (R₁ × R₂) / (R₁ + R₂)
Rₑq = (35 × 20) / (35 + 20)
Rₑq = 700 / 55
Rₑq = 12.73 Ω
Next, we shall determine the total resistance in the circuit. This can be obtained as follow:
Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω
Resistor 3 (R₃) = 15 Ω
Total resistance (R) in the circuit =?
R = Rₑq + R₃ (they are in series connection)
R = 12.73 + 15
R = 27.73 Ω
Finally, we shall determine the current. This can be obtained as follow:
Total resistance (R) = 27.73 Ω
Voltage (V) = 10 V
Current (I) =?
V = IR
10 = I × 27.73
Divide both side by 27.73
I = 10 / 27.73
I = 0.36 A
Therefore, the current is 0.36 A.
Answer:
-0.0789 m
Explanation:
Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:
y = 0.250 sin(-18.4) ≈ -0.0789 m
Notice it is negative since it is below the x-axis.
Answer:
6 A
Explanation:
First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

And now we can use Ohm's law to find the current in the circuit:

Answer:
The correct answer will be "
". The further explanation is given below.
Explanation:
The potential energy will be,
⇒ 
The expression of force will be,
⇒ 
⇒ 
⇒ 
Force seems to be appealing because the expression has been negative. It therefore means that the force or substance is acting laterally in on itself.
Answer:
Conduction is the transfer of heat between substances that are in direct contact with each other. The better the conductor, the more rapidly heat will be transferred. Metal is a good conduction of heat. Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more.