Virtue Ethics (or Virtue Theory) is an approach to Ethics that emphasizes an individual's character as the key element of ethical thinking, rather than rules about the acts themselves (Deontology) or their consequences (Consequentialism).
Just look it up
If two sources emit waves with the same wavelength and a constant phase difference ϕ, they are said to be coherent.
<h3>What is coherent source ?</h3>
- If the frequency and waveform of two wave sources are the same, they are coherent. Waves' optimal quality of coherence makes stationary interference possible.
- When the phase difference between two beams of light is constant, they are coherent; if the phase difference is random or changes, they are noncoherent.
- The concept of a superpositioning at the core of quantum physics and quantum computing is referred to as "quantum coherence." Quantum coherence specifically considers a scenario in which a wave property of an item is split in two and the two waves coherently interfere with one another.
- The interference visibility, which examines the size of the interference fringes in relation to the input waves, is an easy way to measure the degree of coherence; correlation functions provide a precise mathematical definition of the degree of coherence.
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Answer:
The total distance at 7 s is:
![x_{tot}=27\: m](https://tex.z-dn.net/?f=x_%7Btot%7D%3D27%5C%3A%20m)
Explanation:
<u>Distance due to the force</u>
We can use second Newton's law to find the acceleration.
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
![a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D%3D%5Cfrac%7B9%7D%7B5%7D%3D1.8%5C%3A%20m%2Fs%5E%7B2%7D)
Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.
![x_{1}=0.5at_{1}^{2}](https://tex.z-dn.net/?f=x_%7B1%7D%3D0.5at_%7B1%7D%5E%7B2%7D)
![x_{1}=0.5(1.8)(3)^{2}](https://tex.z-dn.net/?f=x_%7B1%7D%3D0.5%281.8%29%283%29%5E%7B2%7D)
![x_{1}=8.1\: m](https://tex.z-dn.net/?f=x_%7B1%7D%3D8.1%5C%3A%20m)
In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.
First, we need to find the final velocity of the first interval
![v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s](https://tex.z-dn.net/?f=v%3Dv_%7Bi%7D%2Bat_%7B1%7D%3D0%2B%281.8%293%3D5.4%5C%3A%20m%2Fs)
So the second distance will be:
![x_{2}=vt_{2}=5.4*4=21.6\: m](https://tex.z-dn.net/?f=x_%7B2%7D%3Dvt_%7B2%7D%3D5.4%2A4%3D21.6%5C%3A%20m)
Therefore, the total distance is:
![x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m](https://tex.z-dn.net/?f=x_%7Btot%7D%3Dx_%7B1%7D%2Bx_%7B2%7D%3D5.4%2B21.6%3D27%5C%3A%20m)
I hope it helps you!
Answer:
a) τ = 4.47746 * 10^25 N-m
b) E = 2.06301 * 10^13 J
c) P = 3.25511*10^21 W
Explanation:
Given:
- The radius of earth r = 6.3781×10^6 m
- The angular speed of earth w = 7.27*10^-5 rad/s
- The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s
- The mass of earth m = 5.972 × 10^24 kg
- The inertia of sphere I = 2/5 * m* r^2
Solution:
- The angular acceleration of the earth from rest to w is given by α:
α = w / t
α = (7.27*10^-5) / (1.57784760 * 10^8)
α = 4.60754*10^-13 rad/s^2
- The required torque τ is given by:
τ = I*α
τ = 2/5 * m* r^2 * α
τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)
τ = 4.47746 * 10^25 N-m
- The power required P to turn the earth to the speed w is:
P = τ*w
P = (4.47746 * 10^25)*(7.27*10^-5)
P = 3.25511*10^21 W
- The energy E required is :
E = P / t
E = (3.25511*10^21) / (1.57784760 * 10^8)
E = 2.06301 * 10^13 J
Answer:
1.11 x 10^6 m
Explanation:
mass of Pluto, M = 1.1 x 10^22 kg
acceleration due to gravity on Pluto, g = 0.59 m/s²
let R be the radius of the Pluto and G is the universal gravitational constant.
G = 6.67 x 10^-11 Nm²/kg²
use the formula for the acceleration due to gravity
![g = \frac{GM}{R^{2}}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7BGM%7D%7BR%5E%7B2%7D%7D)
![R = \sqrt{\frac{GM}{g}}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7Bg%7D%7D)
![R = \sqrt{\frac{6.67\times10^{-11}\times 1.1\times 10^{22}}{0.59}}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B%5Cfrac%7B6.67%5Ctimes10%5E%7B-11%7D%5Ctimes%201.1%5Ctimes%2010%5E%7B22%7D%7D%7B0.59%7D%7D)
R = 1.11 x 10^6 m
Thus, the radius of the Pluto is 1.11 x 10^6 m.