The part of the atom that accounts for electricity is the
2 answers:
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Answer:
The center of mass for the object is from the origin
Explanation:
From the question we are told that
The mass of the first object is
The position of first object with respect to origin
The mass of the second object is
The position of second object with respect to origin
The mass of the third object is
The position of third object with respect to origin
The mass of the fourth object is
The position of fourth object with respect to origin
Generally the center of mass of the object along the x-axis is zero because all the mass lie on the y axis
Generally the location of the center mass of the object is mathematically represented as
=>
=>
Answer:
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω
Explanation:
Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.
When the current increases by 0.450 A wen only R1 is in the circuit, the current is
9/R1 + R2 + 0.450 A = 9/R1 (1)
When the current increases by 0.225 A when only R2 is in the circuit, the current is
9/R1 + R2 + 0.225 A = 9/R2 (2)
equation (1) - (2) equals
9(1/R1 - 1/R2) = 0.450 A - 0.225
9(1/R1 - 1/R2) = 0.125
(1/R1 - 1/R2) = 0.125 A/9 = 0.0138
1/R1 = 0.0138 + 1/R2
R1 = R2/(1 + 0.0138R2) (3)
From (1)
9/R1 - 9/R1 + R2 = 0.450 A
9R2/[R1(R1 + R2)] = 0.450 A
R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5
R2/[R1(R1 + R2)] = 0.5 (4)
From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,
(1 + 0.0138R2) = 0.5(R1 + R2)
0.5R1 + 0.5R2 = 1 + 0.0138R2
0.5R1 = 1 + 0.0138R2 - 0.5R2
0.5R1 = 1 - 0.4862R2 (5)
Substituting (3) into (5) we have
0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2
R2 = (1 + 0.0138R2)(1 - 0.4862R2)
R2 = 1 - 0.4724R2 - 0.0067R2²
Collecting like terms, we have
0.0067R2² + 0.4724R2 + R2 - 1 = 0
0.0067R2² + 1.4724R2 - 1 = 0
Using the quadratic formula,
We choose the positive answer.
So R2 = 0.340 Ω
From (5)
R1 = 0.5 - 0.9931R2
= 0.5 - 0.9931 × 0.340
= 0.5 - 0.338
= 0.162 Ω
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω