<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>
- A certain circuit is composed of two series resistors
- The total resistance is 10 ohms
- One of the resistor is 4 ohms
<h3>
<u>To </u><u>Find </u><u>:</u><u>-</u></h3>
- We have to find the value of other resistor?
<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>
We know that,
In series combination,
- When a number of resistances are connected in series, the equivalent I.e resultant resistance is equal to the sum of the individual resistances and is greater than any individual resistance
<u>That </u><u>is</u><u>, </u>
Rn in series = R1 + R2 + R3.....So on
<u>Therefore</u><u>, </u>
<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>
We have,
R1 + R2 = 10 Ω
4 + R2 = 10Ω
R2 = 10 - 4
R2 = 6Ω
Hence, The value of R2 resistor in series is 6Ω
Answer:
7.74m/s
Explanation:
Mass = 35.9g = 0.0359kg
A = 39.5cm = 0.395m
K = 18.4N/m
At equilibrium position, there's total conservation of energy.
Total energy = kinetic energy + potential energy
Total Energy = K.E + P.E
½KA² = ½mv² + ½kx²
½KA² = ½(mv² + kx²)
KA² = mv² + kx²
Collect like terms
KA² - Kx² = mv²
K(A² - x²) = mv²
V² = k/m (A² - x²)
V = √(K/m (A² - x²) )
note x = ½A
V = √(k/m (A² - (½A)²)
V = √(k/m (A² - A²/4))
Resolve the fraction between A.
V = √(¾. K/m. A² )
V = √(¾ * (18.4/0.0359)*(0.395)²)
V = √(0.75 * 512.53 * 0.156)
V = √(59.966)
V = 7.74m/s
Answer:
-611.32 N/C
0.43723 m
Explanation:
k = Coulomb constant = 
q = Charge = -4.25 nC
r = Distance from particle = 0.25 m
Electric field is given by
The magnitude is 611.32 N/C
The electric field will point straight down as the sign is negative towards the particle.
The distance from the electric field is 1.71436 m
A.) reference group
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