Answer:
T = 2 T₀
Explanation:
To answer this question let's write the expression for electrical conductivity
σ = n e2 τ / m*
The relationship with resistivity is
ρ = 1 /σ
Whereby the resistance
R = ρ L / A = 1 /σ L / A
We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.
As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length
T = 2 T₀
<u><em>Answer:Just as wavelength and frequency are related to light, they are also related to energy. The shorter the wavelengths and higher the frequency corresponds with greater energy. So the longer the wavelengths and lower the frequency results in lower energy.</em></u>
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Explanation:So, if the wavelength of a light wave is shorter, that means that the frequency will be higher because one cycle can pass in a shorter amount of time. ... That means that longer wavelengths have a lower frequency. Conclusion: a longer wavelength means a lower frequency, and a shorter wavelength means a higher frequency!
<em>Extra explanation: All waves can be defined in terms of their frequency and intensity. c = λν expresses the relationship between wavelength and frequency.</em>
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Answer:
Explanation:
la frecuencia = ω/2π, nada cambio
v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima
a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima
la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía
Answer:
200 m\ s Ans .....
Explanation:
Data:
f = 200 Hz
w = 1.0 m
v = ?
Formula:
v = f w
Solution:
v = ( 200)(1.0)
v = 200 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
