Question 1: A mass attached to a spring is initially (t=0) released 5cm away from the equlibrium position. What F0000 should be the initial velocity (v.) in order for the mass to k=18N/m 2kg Vo=? ocillate in 20cm distance? Ó x,=5cm -20cm

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A: Given, A→=6B→=3θA=45°θB=115° Note : According to our guidelines we'll answer the first 3 subparts of…

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A: Given: M=12kg, V=0.0005m3 K) The density of the object is : ρ=MV=120.0005=24000kg/m3

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Q: 13) Explain why a solid undergoes thermal expansion in a microscopic model.

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Q: In the circuit shown, R1=R2=R, and ɛ2 =. The rate of 2 heat dissipated by the whole circuit is 110…

A: R1 = R2 = R  E2 = E1/2  Energy dissipation in the circuit is given by  P = 110 J/s

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A: As per guidelines we are suppose to do only one question from multiple questions please post them…

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A: The potential energy stored in the spring is:

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A: Given data: Period, T=10 s Amplitude, A=5×10-2m

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Q: 14) Themal expansion is a so-called anharmonic process. Why is it called so?

A: Thermal expansion is the tendency of matter to change its shape, area, volume, and density in…

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A: 2) Tension (T) = 910 N number of cycles passing through a point in 0.39 seconds = 91distance between…

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A: Required : The force in p and q coordinates.

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A: Magnitude of electric field is |E| = 5.25×105N/CForce F = kQqr2 = qE . where , q = -6microC

Q: RADIO PARALLEL CIRCUIT V₂ = 9 V R₂ = 60 0 I, = 0.15 A 9 V V₂ = R₂ = 10 02 1₂2=0.9► A CONFIRM V₁= 9 V…

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Q: A 8.82 kg particle with velocity = (5.84 m/s ) - (9.49 m/s ) is at x = 1.64 m, y = 8.10 m. It is…

A: m = 8.82 kg= x = 1.64 my = 8.10 mF = 5.59 N in negative x direction

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