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irina [24]
3 years ago
12

A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen

cy is 0.63 hz. what is the value of m?
Physics
1 answer:
MrMuchimi3 years ago
5 0

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg 


The frequency of mass-spring oscillation is: 
f = (1/2π)√(k/m) 
k = m(2πf)² 

Then we know that k is constant for both trials, we have: 
k = k 


m1(2πf1)² = m2(2πf2)² 

m1 = m2(f2/f1)² 


m1 = (m1+Δm)(f2/f1)² 


m1 = Δm/((f1/f2)²-1)

 m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

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A mover does 422 J of work pushing a crate 8.39 m. How much force did he exert?
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A simple series circuit consists of a 150 Ω resistor, a 29 V battery, a switch, and a 2.1 pF parallel-plate capacitor (initially
Rufina [12.5K]
Find the electric flux and the disp at t=0.50ns 
<span>Given: </span>
<span>Resistor R = 160 Ω </span>
<span>Voltage ε = 22.0 V </span>
<span>Capacitor C = 3.10 pF = 3.10 * 10^-12 F </span>
<span>time t = 0.5 ns = 0.5 * 10^-9 s </span>
<span>ε0 = 8.85 * 10^-12 </span>
<span>Solution: </span>
<span>ELECTRIC FLUX: </span>
<span>Φ = Q/ε0 </span>
<span>we have ε0, we need to find Q the charge </span>
<span>STEP 1: FIND Q </span>
<span>Q = C ε ( 1 - e^(-t/RC) ) </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } </span>
<span>Q = 43.31 * 10^-12 C </span>
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<span>Φ = Q/ε0 </span>
<span>Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } </span>
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<span>DISPLACEMENT CURRENT </span>
<span>we use the following equation: </span>
<span>I = { ε / R } { e^(-t/RC) } </span>
<span>I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>I = { 0.1375 } { 0.365 } </span>
<span>I = 0.0502 A = 0.05 A </span>
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Answer:

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