A 6.0-ohm resistor that obeys Ohm’s Law is connected to a source of variable potential difference. When the applied voltage is d
1 answer:
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Answer:
1. T₁ = 500 N
2. T₂ = 866 N
Explanation:
Please see attached photo for the diagram.
Thus, we can obtain obtained the value of T₁ and T₂ as follow:
1. Determination of T₁
Angle θ = 30
Hypothenus = 100 kg
Opposite = T₁ =?
Sine θ = Opposite /Hypothenus
Sine 30 = T₁ / 100
Cross multiply
T₁ = 100 × Sine 30
T₁ = 100 × 0.5
T₁ = 50 Kg
Multiply by 10 to express in Newton
T₁ = 50 × 10
T₁ = 500 N
2. Determination of T₂
Angle θ = 60
Hypothenus = 100 kg
Opposite = T₂ = ?
Sine θ = Opposite /Hypothenus
Sine 60 = T₂ / 100
Cross multiply
T₂ = 100 × Sine 60
T₂ = 100 × 0.8660
T₂ = 86.6 Kg
Multiply by 10 to express in Newton
T₂ = 86.6 × 10
T₂ = 866 N
Answer:
a.
b.
Explanation:
<u>Given:</u>
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.
At time t = 3 seconds,
<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.
For the time interval of 2 seconds,
The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,
It is the displacement of the particle in 2 seconds.