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4 years ago

1. What is technology?techndegy

Physics

1 answer:

Naya [18.7K]4 years ago

6 0

Answer:

the application of scientific knowledge for practical purposes, especially in industry. Another answer:the sum of techniques,skills,methods and processes used in the production of goods or services or in the accomplishment of objectives, such as scientific investigation

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Wave 1 displaces according the function f(θ)= 2sin θ+1. Wave 2 displaces according to the function f(θ)= 3sin θ+.5. At what angl

r-ruslan [8.4K]

Answer:

θ = 30⁰

Explanation:

We have been given two wave displacement equations,

$y_{1} =2sin\theta +1$ ........(1)

$y_{2}=3sin\theta +0.5$ ........(2)

The waves will have same displacements when $y_{1}=y_{2}=y$ (say)

Therefore, operating (2) - (1), we get,

$sin\theta - 0.5=0$

or, $\theta = sin^{-1}(0.5)=30^{o}$ (since, sin 30⁰ = 0.5)

We can check the answer by putting $\theta=30^{o}$ in equations (1) and (2),

$y_{1}=2sin30^{o}+1=2\times0.5 + 1= 2$

$y_{2}=3sin30^{o}+0.5=3\times0.5+0.5=1.5+0.5=2$

6 0

3 years ago

Read 2 more answers

The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center. If

zvonat [6]

Answer:

The distance between their eye and the ceiling is 33.06 m.

Explanation:

Given that,

The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center, D = 4.35 mm

Diameter of the eye of pupil, d = 5.1 mm

Wavelength, $\lambda=550\ nm$

We need to find the distance between their eye and the ceiling. Using Rayleigh criteria, we get the distance as follows :

$L=\dfrac{Dd}{1.22\lambda}\\\\L=\dfrac{4.35\times 10^{-3}\times 5.1\times 10^{-3}}{1.22\times 550\times 10^{-9}}\\\\L=33.06\ m$

So, the distance between their eye and the ceiling is 33.06 m.

3 0

3 years ago

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto

igor_vitrenko [27]

a) The angular velocity of the electron is $4.12\cdot 10^{16} rad/s$

b) The number of revolutions per second is $6.54\cdot 10^{15}$ rev/s

c) The centripetal acceleration of the electron is $8.98\cdot 10^{22} m/s^2$

Explanation:

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

$\omega = \frac{v}{r}$

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the angular velocity is

$\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s$

The period of revolution of the electron is given by

$T=\frac{2\pi}{\omega}$

where

$\omega = 4.12\cdot 10^{16}rad/s$ is the angular velocity

Substituting,

$T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s$

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

$f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}$ rev/s