Answer:
2.36 μ H
Explanation:
Given,
Number of turns= 90
diameter = 1.3 cm = 0.013 m
unscratched length = 57 cm = 0.57 m
Area, A = π r²
= π x 0.0065² = 1.32 x 10⁻⁴ m²
we know,
![L = \dfrac{\mu_0N^2A}{l}](https://tex.z-dn.net/?f=L%20%3D%20%5Cdfrac%7B%5Cmu_0N%5E2A%7D%7Bl%7D)
![L = \dfrac{4\pi \times 10^{-7}\times 90^2\times 1.32\times 10^{-4}}{0.57}](https://tex.z-dn.net/?f=L%20%3D%20%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2090%5E2%5Ctimes%201.32%5Ctimes%2010%5E%7B-4%7D%7D%7B0.57%7D)
L = 2.36 μ H
Hence, the inductance of the unstretched cord is equal to 2.36 μ H
Answer:
Elastic potential energy, ![E=2.35\times 10^{-8}\ J](https://tex.z-dn.net/?f=E%3D2.35%5Ctimes%2010%5E%7B-8%7D%5C%20J)
Explanation:
Charge, ![q=9.4\times 10^{-10}\ C](https://tex.z-dn.net/?f=q%3D9.4%5Ctimes%2010%5E%7B-10%7D%5C%20C)
Potential, V = 50 V
It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :
![E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B1%7D%7B2%7DqV%5C%5C%5C%5CE%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.4%5Ctimes%2010%5E%7B-10%7D%5Ctimes%2050%5C%5C%5C%5CE%3D2.35%5Ctimes%2010%5E%7B-8%7D%5C%20J)
So, the electric potential energy stored in the capacitor is ![2.35\times 10^{-8}\ J](https://tex.z-dn.net/?f=2.35%5Ctimes%2010%5E%7B-8%7D%5C%20J)
The best type of electromagnetic wave to use to warm food that is safe would be the infrared wave. The correct answer is B.
Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
![d=v_x t](https://tex.z-dn.net/?f=d%3Dv_x%20t)
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
![v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D%3D%5Cfrac%7B3.0%20m%7D%7B1.3%20s%7D%3D2.3%20m%2Fs)
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
![y(t) = h + v_y t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20%2B%20v_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
![0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s](https://tex.z-dn.net/?f=0%3Dh%2Bv_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5Cv_y%20%3D%20%5Cfrac%7B0.5gt%5E2-h%7D%7Bt%7D%3D%5Cfrac%7B0.5%289.8%20m%2Fs%5E2%29%281.3%20s%29%5E2-4.0%20m%7D%7B1.3%20s%7D%3D3.3%20m%2Fs)
So now we can find the magnitude of the initial velocity:
![v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D%3D%5Csqrt%7B%282.3%20m%2Fs%29%5E2%2B%283.3%20m%2Fs%29%5E2%7D%3D4.0%20m%2Fs)