A construction worker is pushing a 50.0-kg box with a force of 150.0 N to the right. If the box is moving at a constant velocity
1 answer:
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Answer:
The correct option is d: T₁ > T₂ > T₃.
Explanation:
Let's evaluate each tension.
<u>Case T₃.</u>
For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.
Since W₃ = mg, where <em>m</em> is for mass and <em>g</em> is for the acceleration due to gravity, we have:
(1) <u>Case T₂.</u>
(2)
By entering W₂ = 2mg and equation (1) into eq (2) we have:
<u>Case T₁.</u>
(3)
Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:
Therefore, the correct option is d: T₁ > T₂ > T₃.
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Answer:
Δe=0.578 kJ/kg
Explanation:
Given data
Velocity v₁=0 m/s
Velocity v₂=34 m/s
to find
Specific energy change Δe
Solution
The specific energy change is simply determined from change in velocity
Δe=(v₂²-v₁²)/2
Put the given values to find the specific energy change
Δe=0.578 kJ/kg
Answer:
12.5 ft/s
Explanation:
Height of person = 6 ft
height of lamp post = 10 ft
According to the question,
dx / dt = 5 ft/s
Let the rate of tip of the shadow moves away is dy/dt.
According to the diagram
10 / y = 6 / (y - x)
10 y - 10 x = 6 y
y = 2.5 x
Differentiate both sides with respect to t.
dy / dt = 2.5 dx / dt
dy / dt = 2.5 (5) = 12.5 ft /s
