Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux
![\phi=NBA](https://tex.z-dn.net/?f=%5Cphi%3DNBA)
Put the value into the formula
![\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2](https://tex.z-dn.net/?f=%5Cphi%3D10%5Ctimes1.0%5Ctimes10%5E%7B-3%7D%5Ctimes%5Cpi%5Ctimes%280.5%5Ctimes10%5E%7B-2%7D%29%5E2)
![\phi=7.85\times10^{-7}\ Tm^2](https://tex.z-dn.net/?f=%5Cphi%3D7.85%5Ctimes10%5E%7B-7%7D%5C%20Tm%5E2)
We need to calculate the induced emf
Using formula of induced emf
![\epsilon=\dfrac{d\phi}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%3D%5Cdfrac%7Bd%5Cphi%7D%7Bdt%7D)
Put the value into the formula
![\epsilon=\dfrac{7.85\times10^{-7}}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7Bdt%7D)
Put the value of emf from ohm's law
![\epsilon =IR](https://tex.z-dn.net/?f=%5Cepsilon%20%3DIR)
![IR=\dfrac{7.85\times10^{-7}}{dt}](https://tex.z-dn.net/?f=IR%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7Bdt%7D)
![Idt=\dfrac{7.85\times10^{-7}}{R}](https://tex.z-dn.net/?f=Idt%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7BR%7D)
![Idt=\dfrac{7.85\times10^{-7}}{0.50}](https://tex.z-dn.net/?f=Idt%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7B0.50%7D)
![Idt=0.00000157=1.57\times10^{-6}\ C](https://tex.z-dn.net/?f=Idt%3D0.00000157%3D1.57%5Ctimes10%5E%7B-6%7D%5C%20C)
We know that,
![Idt=dq](https://tex.z-dn.net/?f=Idt%3Ddq)
![dq=1.57\times10^{-6}\ C](https://tex.z-dn.net/?f=dq%3D1.57%5Ctimes10%5E%7B-6%7D%5C%20C)
We need to calculate the voltage across the capacitor
Using formula of charge
![dq=C dV](https://tex.z-dn.net/?f=dq%3DC%20dV)
![dV=\dfrac{dq}{C}](https://tex.z-dn.net/?f=dV%3D%5Cdfrac%7Bdq%7D%7BC%7D)
Put the value into the formula
![dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}](https://tex.z-dn.net/?f=dV%3D%5Cdfrac%7B1.57%5Ctimes10%5E%7B-6%7D%7D%7B1.0%5Ctimes10%5E%7B-6%7D%7D)
![dV=1.57\ V](https://tex.z-dn.net/?f=dV%3D1.57%5C%20V)
Hence, The voltage across the capacitor is 1.57 V.
Answer:
The average velocity is 0.15 m/s
Explanation:
Use the definition of average velocity as the distance traveled divided the time it took.
Since the movement was on the plane from the origin (0, 0) to the point (-30, 20) corresponding to 30 m west and 20 m north, we calculate the distance using the distance between two points on the plane:
![distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = \sqrt{(-30)^2+20^2} =\sqrt{1300} \approx 36.06\,\,m](https://tex.z-dn.net/?f=distance%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D%20%3D%20%5Csqrt%7B%28-30%29%5E2%2B20%5E2%7D%20%3D%5Csqrt%7B1300%7D%20%5Capprox%2036.06%5C%2C%5C%2Cm)
Then the magnitude of the average velocity can be estimated via the quotient between distance divided time, but since the units required are meters per second, we first convert the four minute time into seconds: 4 * 60 = 240 seconds.
Then the average velocity becomes:
![v_{ave}=\frac{distance}{time} =\frac{36.06}{240} =0.15\,\,m/s](https://tex.z-dn.net/?f=v_%7Bave%7D%3D%5Cfrac%7Bdistance%7D%7Btime%7D%20%3D%5Cfrac%7B36.06%7D%7B240%7D%20%3D0.15%5C%2C%5C%2Cm%2Fs)
Given data:
* The extension of the steel wire is 0.3 mm.
* The length of the wire is 4 m.
* The area of cross section of wire is,
![A=2\times10^{-6}m^2](https://tex.z-dn.net/?f=A%3D2%5Ctimes10%5E%7B-6%7Dm%5E2)
* The young modulus of the steel is,
![Y=2.1\times10^{11}\text{ Pa}](https://tex.z-dn.net/?f=Y%3D2.1%5Ctimes10%5E%7B11%7D%5Ctext%7B%20Pa%7D)
Solution:
The young modulus of the steel in terms of the force and extension is,
![Y=\frac{F\times l}{A\times dl}](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7BF%5Ctimes%20l%7D%7BA%5Ctimes%20dl%7D)
where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,
Substituting the known values,
![\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202.1%5Ctimes10%5E%7B11%7D%3D%5Cfrac%7BF%5Ctimes4%7D%7B2%5Ctimes10%5E%7B-6%7D%5Ctimes0.3%5Ctimes10%5E%7B-3%7D%7D%20%5C%5C%20F%3D0.315%5Ctimes10%5E2%5Ctext%7B%20N%7D%20%5C%5C%20F%3D31.5%5Ctext%7B%20N%7D%20%5Cend%7Bgathered%7D)
Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.
The answer is 86 degrees Fahrenheit. Formula is (30 x 9.5) + 32 = 86
Answer:
a) 2 m/s
b) i) ![K.E = 50 (1.5t^2 + 2) ^2\\](https://tex.z-dn.net/?f=K.E%20%3D%2050%20%281.5t%5E2%20%2B%202%29%20%5E2%5C%5C)
ii) ![F = 3tm](https://tex.z-dn.net/?f=F%20%3D%203tm)
Explanation:
The function for distance is ![x = 0.5t ^3 + 2t](https://tex.z-dn.net/?f=x%20%3D%200.5t%20%5E3%20%2B%202t)
We know that:
Velocity = ![v= \frac{d}{dt} x](https://tex.z-dn.net/?f=v%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%20x)
Acceleration = ![a= \frac{d}{dt}v](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7Bd%7D%7Bdt%7Dv)
To find speed at time t = 0, we derivate the distance function:
![x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2](https://tex.z-dn.net/?f=x%20%3D%200.5%20t%5E3%20%2B%202t%5C%5Cv%3D%20x%27%20%3D%201.5t%5E2%20%2B%202)
Substitute t = 0 in velocity function:
![v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2](https://tex.z-dn.net/?f=v%20%3D%201.5t%5E2%20%2B%202%5C%5Cv%280%29%20%3D%201.5%20%280%29%20%2B%202%5C%5Cv%280%29%20%3D%202)
Velocity at t = 0 will be 2 m/s.
To find the function for Kinetic Energy of the box at any time, t.
![Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2\\](https://tex.z-dn.net/?f=Kinetic%20%5C%20Energy%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%5C%5C%5C%5CK.E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20100%20%5Ctimes%20%281.5t%5E2%20%2B%202%29%20%5E2%5C%5C%5C%5CK.E%20%3D%2050%20%281.5t%5E2%20%2B%202%29%20%5E2%5C%5C)
We know that ![Force = mass \times acceleration](https://tex.z-dn.net/?f=Force%20%3D%20mass%20%5Ctimes%20acceleration)
![a = v'(t) = 1.5t^2 + 2\\a = 3t](https://tex.z-dn.net/?f=a%20%3D%20v%27%28t%29%20%3D%201.5t%5E2%20%2B%202%5C%5Ca%20%3D%203t)
![F = m \times a\\F= m \times 3t\\F = 3tm](https://tex.z-dn.net/?f=F%20%3D%20m%20%5Ctimes%20a%5C%5CF%3D%20m%20%5Ctimes%203t%5C%5CF%20%3D%203tm)