Answer:
s = vcos(x)t
50 = 25cos(45)t
cos(45)t = 2
t = 2/cos(45) = 2sqrt(2)
h = vsin(x)t + gt^2/2
h = 25sin(45)*2sqrt(2) - 4.9*8
h = 10.8 metres
Explanation:
Answer:
The balloon prohibits the flow of air through the air capacitor.
Explanation:
Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.
This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the "mixing paradox".
Answer:
λ = 482.05 nm
Explanation:
The diffraction phenomenon and the diffraction grating is described by the expression
d sin θ = m λ
where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction
in this case they indicate the distance between slits, the angle and the order of diffraction
λ =
d sin θ / m
let's calculate
λ = 1.00 10⁻⁶ sin 74.6 / 2
λ = 4.82048 10⁻⁷ m
Let's reduce to nm
λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)
λ = 482.05 nm
Answer:
a . 0.35cm
b. 11.33cm
Explanation:
a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:
![\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmu_oi_1%7D%7B2%5Cpi%20x%7D%3D%5Cfrac%7B%5Cmu_oi_2%7D%7B2%5Cpi%284-x%29%7D%5C%5C%5C%5C5%2Fx%3D%5Cfrac%7B3.5%7D%7B4-x%7D%5C%5C%5C%5Cx%3D2.35cm%5C%5C%5C%5CN%3D2.35-2%3D0.35cm)
Hence, for currents in same direction, the point is 0.35cm
b. Given both currents flow in opposite directions, the null point lies on the other side.
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:
Let x be distance of N from first wire, then distance from 2nd wire is 4+x:
![\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmu_oi_1%7D%7B2%5Cpi%284%2B%20x%29%7D%3D%5Cfrac%7B%5Cmu_oi_2%7D%7B2%5Cpi%20x%7D%5C%5C%5C%5C5%2F%284%2Bx%29%3D%5Cfrac%7B3.5%7D%7Bx%7D%5C%5C%5C%5Cx%3D9.33cm%5C%5C%5C%5CN%3D9.33%2B2%3D11.33cm)
Hence, if currents are in opposite directions the point on x-axis is 11.33cm