Answer:
The work done is 10.64Joules
Answer:
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
Explanation:
First we calculate the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)
E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 10.35 eV
Now, from Einstein's Photoelectric equation we know that:
Energy of Photon = Work Function + K.E of Electron
10.35 eV = 4.82 eV + K.E
K.E = 10.35 eV - 4.82 eV
<u>K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J</u>
Answer:
One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole. The orbital period of A0620-0090 is 7.75hours, the mass of V616 Monocerotis is estimated to be .67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of the orbit of the orange star.
Explanation:
Observations are used in order to collect data and record a variety of interesting or useful key points about the subject or specimen in observation. These observations, if made well, can be recorded and used to supplement a hypothesis.
Answer:
Gravity is a force of attraction that exists between any two masses, any two bodies, any two particles. Gravity is not just the attraction between objects and the Earth. It is an attraction that exists between all objects, everywhere in the universe.
Explanation:
