A magnetron in a microwave oven emits electromagnetic waves with a frequency f-2450 MHz. What magnetic field strength is require
1 answer:
Answer:
Magnetic field, B = 0.073 T
Explanation:
It is given that,
A magnetron in a microwave oven emits electromagnetic waves with a frequency 2450 MHz,
We need to find the magnetic field required for electrons to move in circular paths with this frequency.
The formula is given by :
q and m are charge and mass of electron
B = 0.073 T
Hence, this is the required solution.
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Answer:
H₀ = 1.6 x 10⁻¹⁸ s⁻¹
Explanation:
The Hubble's Constant can be found by the following formula:
where,
H₀ = Hubble's Constant = ?
v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s
D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)
D = 18.52 x 10²⁴ m
Therefore,
<u>H₀ = 1.6 x 10⁻¹⁸ s⁻¹</u>
Answer:
a) V = 252 cm³
b) Vs = 72 cm³
Explanation:
a)
The volume of the water can be given by the following formula:
<u>V = 252 cm³</u>
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b)
The volume of stone can be given by the change in volume of the water when the stone is dipped into it.
<u>Vs = 72 cm³</u>
Answer:
The maximum speed of the ejected photoelectrons is 1.815 x 10⁶ m/s.
Explanation:
Given;
frequency of the light, f = 3.5 x 10¹⁵ Hz
work function of the metal, Φ = 5.11 eV
Φ = 5.11 x 1.602 x 10⁻¹⁹ J = 8.186 x 10⁻¹⁹ J
The energy of the incident light is given as sum of maximum kinetic energy and work function of the metal.
E = K.E + Φ
where;
E is the energy of the incident light, calculated as;
E = hf
E = (6.626 x 10⁻³⁴)(3.5 x 10¹⁵)
E = 2.319 x 10⁻¹⁸ J
The maximum kinetic energy of the photoelectrons is calculated as;
K.E = E - Φ
K.E = 2.319 x 10⁻¹⁸ J - 8.186 x 10⁻¹⁹ J
K.E = 2.319 x 10⁻¹⁸ J - 0.8186 x 10⁻¹⁸ J
K.E = 1.5004 x 10⁻¹⁸J
The maximum speed of the ejected photoelectrons in 10⁶ m/s is given as;
K.E = ¹/₂mv²
Therefore, the maximum speed of the ejected photon-electrons is 1.815 x 10⁶ m/s.