Answer:
A) E = 4.96 x 10³ eV
B) E = 4.19 x 10⁴ eV
C) E = 3.73 x 10⁹ eV
Explanation:
A)
For photon energy is given as:
![E = hv](https://tex.z-dn.net/?f=E%20%3D%20hv)
![E = \frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where,
E = energy of photon = ?
h = 6.625 x 10⁻³⁴ J.s
λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Therefore,
![E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})](https://tex.z-dn.net/?f=E%20%3D%20%287.95%20x%2010%5E%7B-16%7D%20J%29%28%5Cfrac%7B1%20eV%7D%7B1.6%20x%2010%5E%7B-19%7D%20J%7D%29)
<u>E = 4.96 x 10³ eV</u>
<u></u>
B)
The energy of a particle at rest is given as:
![E = m_{0}c^2](https://tex.z-dn.net/?f=E%20%3D%20m_%7B0%7Dc%5E2)
where,
E = Energy of electron = ?
m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
![E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\](https://tex.z-dn.net/?f=E%20%3D%20%289.1%20x%2010%5E%7B-31%7D%20kg%29%283%20%20x%20%2010%5E8%20m%2Fs%29%5E2%5C%5C)
![E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\](https://tex.z-dn.net/?f=E%20%3D%20%288.19%20x%2010%5E%7B-14%7D%20J%29%28%5Cfrac%7B1%20eV%7D%7B1.6%20x%2010%5E%7B-19%7D%20J%7D%29%5C%5C)
<u>E = 4.19 x 10⁴ eV</u>
<u></u>
C)
The energy of a particle at rest is given as:
![E = m_{0}c^2](https://tex.z-dn.net/?f=E%20%3D%20m_%7B0%7Dc%5E2)
where,
E = Energy of alpha particle = ?
m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
![E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\](https://tex.z-dn.net/?f=E%20%3D%20%286.64%20x%2010%5E%7B-27%7D%20kg%29%283%20%20x%20%2010%5E8%20m%2Fs%29%5E2%5C%5C)
![E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\](https://tex.z-dn.net/?f=E%20%3D%20%285.97%20x%2010%5E%7B-10%7D%20J%29%28%5Cfrac%7B1%20eV%7D%7B1.6%20x%2010%5E%7B-19%7D%20J%7D%29%5C%5C)
<u>E = 3.73 x 10⁹ eV</u>
The buoyant force exerted on a 6,000-mL toy balloon by the air surrounding the balloon is 52.97N.
<h3>How to calculate buoyant force?</h3>
The buoyancy needed for an object can be calculated using the formula;
B = ρ × V × g
where;
- ρ and V are the object's density and volume respectively
- g is the acceleration due to gravity (9.81m/s²)
B = 6000 × 0.0009 × 9.81
B = 52.97N
Therefore, the buoyant force exerted on a 6,000-mL toy balloon by the air surrounding the balloon is 52.97N.
Learn more about buoyant force at: brainly.com/question/21990136
#SPJ1
Answer: C. Acceleration
Explanation:
Acceleration
is defined as the variation of Velocity
in time
:
In addition, acceleration is a vector quantity (like velocity) and as a vctor it has magnitude and direction. <u>When one of this aspects changes (or both) acceleration changes as well.</u>
So, if the car keeps the same speed (
) but changes its direction, its acceleration changes.
Answer:
Chicken and peanut butter lol
Explanation:
Answer:
The position of the image from the camera lens is approximately 121.1 mm
Explanation:
The given parameters of the lens are;
The specification of the camera lens = 120 mm
Therefore, the focal length of the camera lens, f = 120 mm = 0.12 m
The distance of the object from the camera,
= 13 m
The lens equation for finding the position of the image is given as follows;
![\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%20%3D%20%5Cdfrac%7B1%7D%7Bd_o%7D%20%2B%20%5Cdfrac%7B1%7D%7Bd_i%7D)
Where;
= The position of the image from the camera lens
Therefore, by plugging in the known values, we have;
![\dfrac{1}{13} = \dfrac{1}{0.12} + \dfrac{1}{d_i}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B13%7D%20%3D%20%5Cdfrac%7B1%7D%7B0.12%7D%20%2B%20%5Cdfrac%7B1%7D%7Bd_i%7D)
![\dfrac{1}{d_i} = \dfrac{1}{0.12} - \dfrac{1}{13} = \dfrac{13 - 0.12}{0.12 \times 13} = \dfrac{12.88}{1.56} = \dfrac{322}{39}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bd_i%7D%20%20%3D%20%5Cdfrac%7B1%7D%7B0.12%7D%20-%20%20%5Cdfrac%7B1%7D%7B13%7D%20%3D%20%5Cdfrac%7B13%20-%200.12%7D%7B0.12%20%5Ctimes%2013%7D%20%3D%20%5Cdfrac%7B12.88%7D%7B1.56%7D%20%3D%20%5Cdfrac%7B322%7D%7B39%7D)
![\therefore d_i = \dfrac{39}{322} \approx 0.1211](https://tex.z-dn.net/?f=%5Ctherefore%20d_i%20%3D%20%5Cdfrac%7B39%7D%7B322%7D%20%20%5Capprox%20%200.1211)
The position of the image from the camera lens,
= 0.1211 m = 121.1 mm