A 1.695 kg block of lead is 10 cm long, 5 cm high, and 30mm wide. Calculate the density of the lead
1 answer:
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Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
Answer:
The answer is B.
Explanation:
Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :
Total resistance = 20Ω + 40Ω + 60Ω
= 120Ω
Next, we have to find out that 1Ω is equal to how many voltage by dividing :
120Ω = 60V
1Ω = 60V ÷ 120
1Ω = 0.5V
Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :
1Ω = 0.5V
20Ω = 0.5V × 20
20Ω = 10V
