Answer:
1.) A simple harmonic oscillator has an amplitude of 3.50 cm and a maximum speed of 26.0 cm/s. What is its speed when the displacement is 1.75 cm? 2.) Both pendulum A and B are 3.0 m long. The period of A is T. Pendulum A is twice as heavy as pendulum B. What is the period of B? 3.) The time for one cycle of a periodic process is called the _ ? 4.) In simple harmonic motion, the acceleration is proportional to? 5.) The position of a mass that is oscillating on a spring is given by x= (18.3 cm) cos [(2.35 s-1)t]. What is the frequency of this motion?
Explanation:
Mass of solid cube = 725g = 0.725kg Length of each side = 6.35cm =0.0635 m As volume = length ×length×length Volume= 0.0635×0.0635×0.0635 =0.1905 m³ ρ = m/V ρ= 0.725/0.1905 = 0.138 kg/m³ Answer
Answer:
![E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)](https://tex.z-dn.net/?f=E_%7B2%7D%20%28t%29%20%3D%20-%5Cpi%5Cmu_o%7D%20%5Crho%5E%7B2%7D%20n_%7B1%7Dn_%7B2%7DL%5Cfrac%7Bd%20%7D%7Bdt%7DI_%7B1%7D%28t%29)
Explanation:
Consider two solenoids that are wound on a common cylinder as shown in fig. 1. Let the cylinder have radius 'ρ' and length 'L' .
No. of turns of solenoid 1 = n₁
No. of turns of solenoid 1 = n₂
Assume that length of solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length:
![\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I](https://tex.z-dn.net/?f=%5Coint%20%5Coverrightarrow%20%7BB%7D%5Coverrightarrow%20%7B%28r%29%7D.%5Coverrightarrow%20%7Bdl%7D%3D%20%5Cmu_%7Bo%7DI)
We will consider the field that arises from solenoid 1, having n₁ turns per unit length. The magnetic field due to solenoid 1 passes through solenoid 2, which has n₂ turns per unit length.
Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction:
![\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)](https://tex.z-dn.net/?f=%5Coint%20%5Coverrightarrow%20%7BB%7D%5Coverrightarrow%20%7B%28r%29%7D.%5Coverrightarrow%20%7Bdl%7D%3D%20-%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cphi%20_%7BM%7D%28t%29)
Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)
Using:
--- (2)
Flux generated due to magnetic field B₁
---(3)
area of solenoid = ![A = \pi \rho^{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20%5Crho%5E%7B2%7D)
substituting (2) in (3)
----(4)
We have to find electromotive force E₂(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1, i.e.
---- (5)
substituting (4) in (5)
![E_{2} (t) = -n_{2}L\frac{d }{dt}(\mu_o} \pi \rho^{2} n_{1}I_{1}(t))\\E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)](https://tex.z-dn.net/?f=E_%7B2%7D%20%28t%29%20%3D%20-n_%7B2%7DL%5Cfrac%7Bd%20%7D%7Bdt%7D%28%5Cmu_o%7D%20%5Cpi%20%5Crho%5E%7B2%7D%20n_%7B1%7DI_%7B1%7D%28t%29%29%5C%5CE_%7B2%7D%20%28t%29%20%3D%20-%5Cpi%5Cmu_o%7D%20%5Crho%5E%7B2%7D%20n_%7B1%7Dn_%7B2%7DL%5Cfrac%7Bd%20%7D%7Bdt%7DI_%7B1%7D%28t%29)
C.Earth is the largest and most dense of the terrestrial planets
To solve this problem we will apply the concept related to the Poisson ratio for which the longitudinal strains are related, versus the transversal strains. First we need to calculate the longitudinal strain as follows
![\epsilon_x = \frac{l_f-l_i}{l_i}](https://tex.z-dn.net/?f=%5Cepsilon_x%20%3D%20%5Cfrac%7Bl_f-l_i%7D%7Bl_i%7D)
![\epsilon_x = \frac{(9.80554)-(9.8)}{9.8}](https://tex.z-dn.net/?f=%5Cepsilon_x%20%3D%20%5Cfrac%7B%289.80554%29-%289.8%29%7D%7B9.8%7D)
![\epsilon_x = 0.0005653](https://tex.z-dn.net/?f=%5Cepsilon_x%20%3D%200.0005653)
Second we will calculate the lateral strain as follows
![\epsilon_y = \frac{a_f-a_i}{a_i}](https://tex.z-dn.net/?f=%5Cepsilon_y%20%3D%20%5Cfrac%7Ba_f-a_i%7D%7Ba_i%7D)
![\epsilon_y = \frac{2.59952-2.6}{2.6}](https://tex.z-dn.net/?f=%5Cepsilon_y%20%3D%20%5Cfrac%7B2.59952-2.6%7D%7B2.6%7D)
![\epsilon_y = -0.0001846153](https://tex.z-dn.net/?f=%5Cepsilon_y%20%3D%20-0.0001846153)
The Poisson's ratio is the relation between the two previous strain, then,
![\upsilon = -\frac{\epsilon_y}{\epsilon_x}](https://tex.z-dn.net/?f=%5Cupsilon%20%3D%20-%5Cfrac%7B%5Cepsilon_y%7D%7B%5Cepsilon_x%7D)
![\upsilon = -\frac{(-0.0001846153)}{0.0005653}](https://tex.z-dn.net/?f=%5Cupsilon%20%3D%20%20-%5Cfrac%7B%28-0.0001846153%29%7D%7B0.0005653%7D)
![\upsilon = 0.3265](https://tex.z-dn.net/?f=%5Cupsilon%20%3D%200.3265)
Therefore the Poisson's ratio for the material is 0.3265