1) metal
Even though metalloids are also conductors of heat and electricity, malleable they are not as good as metals.
Metals are very good conductors of electricity and heat. They are also very hard to touch. Noble gases and non metals are the exact opposite in physical and chemical properties. Metals readily react with oxygen.
Answer:
80% (Eighty percent)
Explanation:
The material has a refractive index (n) of 1.25
Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s
We can find the speed of light in the material (v) using the relationship
n = c/v, similarly
v = c/n
therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s
v = 239 833 966 m/s
Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as
(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%
Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:
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Answer:
Explanation:
The mass of the deuteron = mass of the proton + mass of the neutron + mass equivalent of the energy of 2.2 Mev evolved.
I amu = 931 Mev
2.2 Mev = 2.2 / 931 amu
= ( 2.2 / 931 )x 1.6726 x 10⁻²⁷
= .00395 x 10⁻²⁷
The mass of the deuteron =( 1.6726 + 1.6749 + .00395)x 10⁻²⁷ kg
= 3.35145 x 10⁻²⁷ kg
b ) Momentum of gamma ray
= h / λ ( h is plank's constant and λ is wavelength of gamma ray )
= hυ / υλ ( υ is frequency of gamma ray )
= E / c ( E is energy of photon and c is velocity o light )
= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹ J / 3 x 10⁸
= 1.173 x 10⁻²¹ Kg m /s
This will be the momentum of deuteron also
Kinetic energy
= p² / 2m ( p is momentum and m is mass of deuteron )
= ( 1.173 x 10⁻²¹ )² / ( 2 x 3.35145 x 10⁻²⁷)
= 1.376 x ⁻¹⁵ J
Energy of gamma ray
= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 3.52 x 10⁻¹³ J
So kinetic energy of deuteron is smaller than energy of gamma ray photon .
The frequency of the radio wave is:

The wavelength of an electromagnetic wave is related to its frequency by the relationship

where c is the speed of light and f the frequency. Plugging numbers into the equation, we find

and this is the wavelength of the radio waves in the problem.