The claim that can be deduced is that B. Warmer weather caused Galetown to have more severe rainstorms.
<h3>What is a claim?</h3>
It should be noted that a claim simply means the stance of an author in a literary work.
In this case, the claim is that warmer weather caused Galetown to have more severe rainstorms.
When the temperature of an air parcel begins higher, this will make it rise higher in the troposphere before the temperature of the air parcel and surrounding air are equal.
Learn more about claim on:
brainly.com/question/2748145
1) First of all, let's find the resistance of the wire by using Ohm's law:
![V=IR](https://tex.z-dn.net/?f=V%3DIR)
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
![R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega](https://tex.z-dn.net/?f=R%3D%20%5Cfrac%7BV%7D%7BI%7D%3D%20%5Cfrac%7B5.70%20V%7D%7B17.6%20A%7D%3D0.32%20%5COmega%20%20)
2) Now that we have the value of the resistance, we can find the resistivity of the wire
![\rho](https://tex.z-dn.net/?f=%5Crho)
by using the following relationship:
![\rho = \frac{RA}{L}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%20%5Cfrac%7BRA%7D%7BL%7D%20)
Where A is the cross-sectional area of the wire and L its length.
We already have its length
![L=2.90 m](https://tex.z-dn.net/?f=L%3D2.90%20m)
, while we need to calculate the area A starting from the radius:
![A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%280.654%5Ccdot%2010%5E%7B-3%7Dm%29%5E2%3D1.34%20%5Ccdot%2010%5E%7B-6%7Dm%5E2)
And now we can find the resistivity:
To solve this problem it is necessary to apply the concepts related to Period in a spring, elastic potential energy and energy in simple harmonic movement.
From the definition we know that the period can be expressed as
![T = 2\pi \sqrt{\frac{m}{k}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D)
Where
m= Mass
k = Spring constant
Our values are given as
m = 16.6Kg
K = 2470N/m
Therefore the period would be
![T = 2\pi \sqrt{\frac{m}{k}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7D)
![T = 2\pi \sqrt{\frac{16.6}{2470}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7B16.6%7D%7B2470%7D%7D)
![T = 0.515s](https://tex.z-dn.net/?f=%20T%20%3D%200.515s%20)
PART A) The frequency is the inverse of the period therefore
![f = \frac{1}{T}\\f = \frac{1}{0.515}\\f = 1.9417Hz](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7BT%7D%5C%5Cf%20%3D%20%5Cfrac%7B1%7D%7B0.515%7D%5C%5Cf%20%3D%201.9417Hz)
PART B) Elastic potential energy depends on compression and elasticity therefore
![PE = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
![PE = \frac{1}{2} (2470)(0.281)^2](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%282470%29%280.281%29%5E2)
![PE = 97.51J](https://tex.z-dn.net/?f=PE%20%3D%2097.51J)
PART C) Kinetic energy depends on mass and speed therefore
![KE = \frac{1}{2}mv^2\\KE = \frac{1}{2} (16.6)(12.6)^2\\KE = 1371.7J](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5CKE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2816.6%29%2812.6%29%5E2%5C%5CKE%20%3D%201371.7J)
PART D) As energy is conserved, the total energy is equivalent to the dual ratio of the elasticity constant and the amplitude, mathematically,
![KE+PE = \frac{1}{2}kA^2](https://tex.z-dn.net/?f=KE%2BPE%20%3D%20%5Cfrac%7B1%7D%7B2%7DkA%5E2)
![1371.7+97.51 = \frac{1}{2}2470A^2](https://tex.z-dn.net/?f=1371.7%2B97.51%20%3D%20%5Cfrac%7B1%7D%7B2%7D2470A%5E2)
![A = \sqrt{\frac{1415*2}{2470}}](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%5Cfrac%7B1415%2A2%7D%7B2470%7D%7D)
![A = 1.07m](https://tex.z-dn.net/?f=A%20%3D%201.07m)
Therefore the motion's amplitude is 1.07m
Answer:
(a) 4.27 x 10^-4 Telsa
(b) 3.28 x 10^-4 Telsa
Explanation:
side of square, a = 5.49 cm
inner radius, r = 18.1 cm = 0.181 m
number of turns,N = 450
current, i = 0.859 A
(a)
The magnetic field due to a solenoid due to inner radius is
![B = \frac{\mu_{0}Ni}{2\pi r_{inner}}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_%7B0%7DNi%7D%7B2%5Cpi%20r_%7Binner%7D%7D)
![B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.181}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B4%5Cpi%5Ctimes%2010%5E%7B-7%7D%5Ctimes%20450%5Ctimes%200.859%7D%7B2%5Cpi%20%5Ctimes%200.181%7D)
B = 4.27 x 10^-4 Telsa
(b)
The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m
The magnetic field due to the outer radius is
![B = \frac{\mu_{0}Ni}{2\pi r_{outer}}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_%7B0%7DNi%7D%7B2%5Cpi%20r_%7Bouter%7D%7D)
![B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.236}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B4%5Cpi%5Ctimes%2010%5E%7B-7%7D%5Ctimes%20450%5Ctimes%200.859%7D%7B2%5Cpi%20%5Ctimes%200.236%7D)
B = 3.28 x 10^-4 Tesla
The resistance of the silver block is given by
![R= \frac{\rho L}{A}](https://tex.z-dn.net/?f=R%3D%20%5Cfrac%7B%5Crho%20L%7D%7BA%7D%20)
where
![\rho=1.63 \cdot 10^{-6} \Omega \cdot cm](https://tex.z-dn.net/?f=%5Crho%3D1.63%20%5Ccdot%2010%5E%7B-6%7D%20%5COmega%20%5Ccdot%20cm)
is the silver resistivity
![L=10 cm](https://tex.z-dn.net/?f=L%3D10%20cm)
is the length of the block
![A=0.10 cm^2](https://tex.z-dn.net/?f=A%3D0.10%20cm%5E2)
is the cross-sectional area of the block
If we plug the data into the equation, we find the resistance of the silver block: