The radius, r, of the child from the center of the wheel is
r = 1.3 m
The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s
The linear speed of the child is the tangential velocity, given by
v = rω
= (1.3 m)*(1.496 rad/s)
= 1.945 m/s
Answer: 1.95 m/s (nearest hundredth)
Answer:
R2 = 300 Ohms
Explanation:
Let the two resistors be R1 and R2 respectively.
RT is the total equivalent resistance.
Given the following data;
R1 = 100 Ohms
RT = 75 Ohms
To find R2;
Mathematically, the total equivalent resistance of resistors connected in parallel is given by the formula;

Substituting into the formula, we have;

Cross-multiplying, we have;
75 * (100 + R2) = 100R2
7500 + 75R2 = 100R2
7500 = 100R2 - 75R2
7500 = 25R2
R2 = 7500/25
R2 = 300 Ohms
Answer: c
Explanation:
The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.
a. 2×2 atoms of Al ≠ 3×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!
b. 3×2 atoms of Al ≠ 3×1 atoms of Al
3×3 atoms of O ≠ 2×2 atoms of O
c. 2×2 atoms of Al = 4×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
BOTH ARE EQUAL, CORRECT ANSWER!!!
d. 2×2 atoms of Al ≠ 1×1 atoms of Al
2×3 atoms of O = 3×2 atoms of O
Answer:
The mass of the other worker is 45 kg
Explanation:
The given parameters are;
The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker
The mass of the lighter construction worker, m₁ = 90 kg
The height level of the lighter construction worker's location = h₁
The height level of the other construction worker's location = h₂ = 2·h₁
The gravitational potential energy, P.E., is given as follows;
P.E. = m·g·h
Where;
m = The mass of the object at height
g = The acceleration due to gravity
h = The height at which is located
Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker
We have;
P.E.₁ = P.E.₂
Therefore;
m₁·g·h₁ = m₂·g·h₂
h₂ = 2·h₁
We have;
m₁·g·h₁ = m₂·g·2·h₁
m₁ = 2·m₂
90 kg = 2 × m₂
m₂ = (90 kg)/2 = 45 kg
The mass of the other construction worker is 45 kg.
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