Answer
Wavelength= 30*20^8/30=10^7m
Explanation:
Velocity = frequency *wavelength
We're frequency=30HZ
Velocity of light= 3*10^8m/s
Wavelength= 30*20^8/30=10^7m
Explanation:
a simple lifting machine consisting of a rope which unwinds from a wheel on to a cylindrical drum or shaft joined to the wheel to provide mechanical advantage. reeeeeeeeeeeeeeeeeeeeeeeeee
(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.
(b) The speed of the proton is determined as 3.1 x 10⁹ m/s.
<h3>
Momentum of the proton</h3>
The momentum of the proton is calculated as follows;
K.E = ¹/₂mv²
where;
- m is mass of proton = 1.67 x 10⁻²⁷ kg
- v is speed of the proton = ?
<h3>Speed of the proton</h3>
v² = 2K.E/m
v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)
v² = 9.6 x 10¹⁸
v = 3.1 x 10⁹ m/s
<h3>Momentum of the proton</h3>
P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s
Learn more about momentum here: brainly.com/question/7538238
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Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A =
= 6.77 ×
m²
Now Area of cylinder is :
A =
d²
solving for d:
d = 
d = 9.28 cm
Answer:

Explanation:
From the question we are told that:
Frequency of 3rd harmonics 
Frequency of 5th harmonics 
Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

Therefore

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

Therefore

Generally the equation for the frequency of the first harmonic is mathematically given by


