Answer:
the required diameter of the rod is 9.77 mm
Explanation:
Given:
Length = 1.5 m
Tension(P) = 3 kN = 3 × 10³ N
Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa
E = 70 GPa = 70 × 10⁹ Pa
δ = 1 mm = 1 × 10⁻³ m
The required diameter(d) = ?
a) for stress
The stress equation is given by:
A is the area = πd²/4 = (3.14 × d²)/4
Substituting the values, we get
d = (9.77 × 10⁻³) m
d = 9.77 mm
b) for deformation
δ = (P×L) / (A×E)
A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063
d² = (4 × A) / π = (0.000063 × 4) / 3.14
d² = 0.0000819
d = 9.05 × 10⁻³ m = 9.05 mm
We use the larger value of diameter = 9.77 mm
Answer:
8 μC
Explanation:
By definition, current is the rate of change of charge, so we can write the following equation for current I:
I = ΔQ / Δt
As charge must be conserved, all the charge carried by the current must add to the charge on the plates of the capacitor, so we can finf this incremental charge as follows:
ΔQ = I* Δt (assuming that current remains constant during the charging process)
⇒ ΔQ = 3 A* 2 μsec = 3 coul/sec*2 μsec = 6 μC
As the initial charge must be conserved also, the magnitude of the net electric charge of the capacitor must be as follows:
Qnet = Q₀+ ΔQ = 2 μC + 6 μC = 8 μC
Answer:
Explanation:
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