Answer:
m = 0.59 kg.
Explanation:
First, we need to find the relation between the frequency and mass on a spring.
The Hooke's law states that
![F = -kx](https://tex.z-dn.net/?f=F%20%3D%20-kx)
And Newton's Second Law also states that
![F = ma = m\frac{d^2x}{dt^2}](https://tex.z-dn.net/?f=F%20%3D%20ma%20%3D%20m%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D)
Combining two equations yields
![a = -\frac{k}{m}x](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7Bk%7D%7Bm%7Dx)
The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.
![\omega = \sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
And given that ω = 2πf
the relation between frequency and mass becomes
.
Let's apply this to the variables in the question.
![0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg](https://tex.z-dn.net/?f=0.88%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%5C%5C0.60%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%2B0.68%7D%7D%5C%5C%5Cfrac%7B0.88%7D%7B0.60%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D%7D%7B%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%2B0.68%7D%7D%7D%5C%5C1.4667%20%3D%20%5Cfrac%7B%5Csqrt%7Bm%2B0.68%7D%7D%7B%5Csqrt%7Bm%7D%7D%5C%5C2.15m%20%3D%20m%20%2B%200.68%5C%5C1.15m%20%3D%200.68%5C%5Cm%20%3D%200.59~kg)
The correct answer is option C. insulator.
An insulator material<span> has a high resistivity and prevents the movement of electrons. It is a hindrance in the path of electrons & resists the current flow.
The conductor conducts electrons as it offers very less resistance. A battery provides voltage to the circuit.
Hence, option (c) is correct. </span>
Answer:
Electric field will be ![2.33\times 10^{-2}V/m](https://tex.z-dn.net/?f=2.33%5Ctimes%2010%5E%7B-2%7DV%2Fm)
Explanation:
We have given diameter of the copper d = 2.05 mm
So radius ![r=\frac{d}{2}=\frac{2.05}{2}=1.05mm=1.05\times 10^{-3}m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B2.05%7D%7B2%7D%3D1.05mm%3D1.05%5Ctimes%2010%5E%7B-3%7Dm)
We know that area ![A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%3D3.14%5Ctimes%20%281.05%5Ctimes%2010%5E%7B-3%7D%29%5E2%3D3.461%5Ctimes%2010%5E%7B-6%7Dm%5E2)
Current is given as i = 4.7 A
So current density ![j=\frac{i}{A}=\frac{4.70}{3.461\times 10^{-6}}=1.357\times 10^6A/m^2](https://tex.z-dn.net/?f=j%3D%5Cfrac%7Bi%7D%7BA%7D%3D%5Cfrac%7B4.70%7D%7B3.461%5Ctimes%2010%5E%7B-6%7D%7D%3D1.357%5Ctimes%2010%5E6A%2Fm%5E2)
Resistivity is given ![\rho =1.72\times 10^{-8}ohm-m](https://tex.z-dn.net/?f=%5Crho%20%3D1.72%5Ctimes%2010%5E%7B-8%7Dohm-m)
We know that electric field is given by ![E=\rho j=1.72\times 10^{-8}\times 1.357\times 10^6=2.33\times 10^{-2}V/m](https://tex.z-dn.net/?f=E%3D%5Crho%20j%3D1.72%5Ctimes%2010%5E%7B-8%7D%5Ctimes%201.357%5Ctimes%2010%5E6%3D2.33%5Ctimes%2010%5E%7B-2%7DV%2Fm)
Organ, integument, dermis
Answer:
0.365 cm³
Explanation:
The change in volume is found by multiplying the coefficient of expansion by the volume and the temperature change. The temperature change is in °F, but the expansion coefficient is per °C, so we need to convert the temperature scale in the computation.
ΔV = V·Ce·ΔT
= (π/6·d³)(1.5×10⁻⁶/°C)((5 °C)/(9 °F))(305 °F)
= (1436.76 cm³)(1.5×10⁻⁶/°C)(169.44 °C)
= 0.365 cm³ . . . . increase in volume