Answer:
Sempre que os corpos vibram e as vibrações se propagam num meio material como o ar, produzem ondas sonoras que são ondas de pressão?
Explanation:
Answer:
speed = 1.24 × 10⁸m/s
frequency = 4.74 × 10¹⁴Hz
wavelength = 262nm
Explanation:
the speed of the helium-neon light in zircon is given by,
v = c / n
c = 3 × 10⁸m/s is the speed of light in vacuum (and in air)
n = 2.419 is the refractive index of diamond
v = 3 × 10⁸ / 2.419
= 1.24 × 10⁸m/s
(b) Frequency
The wavelength of the light in air is:
λ₀ = 632.8 × 10⁻⁹
The frequency of the light does not depend on the medium, so it is equal in air and in diamond. Therefore, we can calculate the frequency by using the speed of light in air and the wavelength in air:
f₀ = c / λ₀
= 3 × 10⁸ / 632.8 × 10⁻⁹
= 4.74 × 10¹⁴Hz
and the frequency of the light indiamond is the same:
f¹ = f₀ = 4.74 × 10¹⁴Hz
(c) Wavelength
To calculate the wavelength of the light in daimond, we can use the relationship between speed of light in diamond and frequency:
λ¹ = v / f¹
= 1.24 × 10⁸ / 4.74 × 10¹⁴
= 2.62 × 10⁻⁷m
= 262nm
Answer:
No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be 
times larger.
Explanation:
The work done on the ball is equal to the kinetic energy gained by the ball:
So when the work done doubles, the kinetic energy doubles as well:
However, the kinetic energy is given by
where
m is the mass of the ball
v is its speed
We see that the kinetic energy is proportional to the square of the speed, 
. We can rewrite the last equation as
which also means
If the work is doubled,
So the new speed is
So, the speed is 
times larger.
Answer:
b. Light ➪ mechanical ➪ electrical
<h2>Answer</h2>
Option A that is 8.8 × 10^3 m/s
<h2>Explanation</h2>
The magnetic field B is defined from the Lorentz Force Law, and specifically from the magnetic force on a moving charge. It says
Field-strength = BVqsinΔ
<h2>v = E/B </h2>
Since field are perpendicular so sin90 = 1
v = 4.6/10^4 / 5.2
v = 8846.15 m /s
The speed at which electrons pass through the selector without deflection = 8846.15 m /s
