We are given with
W = <span>2.34 J
D = 3.4 cm
L = 4.6 cm
To get the pressure of the steam we use the formula
W = P</span>ΔV
2.36 J = P π (3.4 x 10-2)²/4 (4.6 x 10-2)
P = 56508 Pa
<h3><u>Answer;</u></h3>
D. An image that is smaller than the object and is behind the mirror
<h3><u>Explanation;</u></h3>
- Magnification is the ratio of the height of the image to the height of the object.
- When magnification, m, is greater than 1 the image is larger than the object, and when m is less than 1 means the image is smaller than the object.
- If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.
- Magnification is negative when the image is inverted, therefore a real image.
To solve this problem it is necessary to use the concepts related to Snell's law.
Snell's law establishes that reflection is subject to
![n_1sin\theta_1 = n_2sin\theta_2](https://tex.z-dn.net/?f=n_1sin%5Ctheta_1%20%3D%20n_2sin%5Ctheta_2)
Where,
Angle between the normal surface at the point of contact
n = Indices of refraction for corresponding media
The total internal reflection would then be given by
![n_1 sin\theta_1 = n_2sin\theta_2](https://tex.z-dn.net/?f=n_1%20sin%5Ctheta_1%20%3D%20n_2sin%5Ctheta_2)
![(1.54) sin\theta_1 = (1.33)sin(90)](https://tex.z-dn.net/?f=%281.54%29%20sin%5Ctheta_1%20%3D%20%281.33%29sin%2890%29)
![sin\theta_1 = \frac{1.33}{1.54}](https://tex.z-dn.net/?f=sin%5Ctheta_1%20%3D%20%5Cfrac%7B1.33%7D%7B1.54%7D)
![\theta = sin^{-1}(\frac{1.33}{1.54})](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20sin%5E%7B-1%7D%28%5Cfrac%7B1.33%7D%7B1.54%7D%29)
![\theta = 59.72\°](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2059.72%5C%C2%B0)
Therefore the
would be equal to
![\alpha = 90\°-\theta](https://tex.z-dn.net/?f=%5Calpha%20%3D%2090%5C%C2%B0-%5Ctheta)
![\alpha = 90-59.72](https://tex.z-dn.net/?f=%5Calpha%20%3D%2090-59.72)
![\alpha = 30.27\°](https://tex.z-dn.net/?f=%5Calpha%20%3D%2030.27%5C%C2%B0)
Therefore the largest value of the angle α is 30.27°
Answer:
the new current on the wire is 3.64 A.
Explanation:
Given;
first force on the wire, F₁ = 0.026 N
second force on the wire, F₂ = 0.063 N
first current on the wire, I₁ = 1.5 A
second current on the wire, I₂ = ?
The force on a current carrying conductor placed in a magnetic field is given as;
![F = BIL(sin \theta)\\\\](https://tex.z-dn.net/?f=F%20%3D%20BIL%28sin%20%5Ctheta%29%5C%5C%5C%5C)
F ∝ I
![\frac{F_1}{I_1} = \frac{F_2}{I_2} \\\\I_2 = \frac{F_2I_1}{F_1} \\\\I_2 = \frac{0.063\ \times\ 1.5 }{0.026} \\\\I_2 = 3.64 \ A](https://tex.z-dn.net/?f=%5Cfrac%7BF_1%7D%7BI_1%7D%20%3D%20%5Cfrac%7BF_2%7D%7BI_2%7D%20%5C%5C%5C%5CI_2%20%3D%20%5Cfrac%7BF_2I_1%7D%7BF_1%7D%20%5C%5C%5C%5CI_2%20%3D%20%5Cfrac%7B0.063%5C%20%5Ctimes%5C%201.5%20%7D%7B0.026%7D%20%5C%5C%5C%5CI_2%20%3D%203.64%20%5C%20A)
Therefore, the new current on the wire is 3.64 A.