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3 years ago

How did Galileo increase public support for Copernicus’s model?

Physics

2 answers:

miskamm [114]3 years ago

8 0

I think by using data collected by Tycho Brahe

just olya [345]3 years ago

3 0

Answer:

Explanation:

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A medical ultrasound imaging system sends out a steady stream of very short pulses. To simplify analysis, the reflection of one

arlik [135]

Answer:

Minimum time = 1.95x10^-4 s

Number of pulses = 5128.21 pulses/s

Explanation:

We have the speed of sound waves through human tissue with a value of 1540 m/s, to calculate the time it takes for the pulse to travel a distance of 30 cm (since the pulse will first travel a distance of 15 cm and then it will return another 15 cm to be detected by the equipment), therefore, the time between the two pulses will be equal to:

tminimum = 0.30 m/1540 m/s = 1.95x10^-4 s

To calculate the number of pulses, one second must be divided over the minimum time between the two pulses, as follows:

npulses = 1 s/1.95x10^-4 s = 5128.21 pulses/s

5 0

3 years ago

In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is 1.0 cm. What happens

kobusy [5.1K]

Answer:

C) It increases to 2.0 cm

Explanation:

In a double-slit diffraction experiment, the distance on the screen between two adjacent maxima is given by

$\Delta y = \frac{\lambda D}{d}$

where

$\lambda$ is the wavelength of the wave

D is the distance of the screen from the slits

d is the separation between the slits

In this problem, the initial distance between adjacent maxima is 1.0 cm. Later, the slit separation is cut in a half, which means that the new slit separation is

$d'=\frac{d}{2}$

Substituting into the equation, we find that the new separation between the maxima is

$\Delta y' = \frac{\lambda D}{d/2}=2(\frac{\lambda D}{d})=2\Delta y$

So, the distance increases by a factor 2: therefore, the new separation between the maxima will be 2.0 cm.

5 0

3 years ago

When did the Space Age begin?

Marta_Voda [28]

The Space Age began on October 4, 1957.

4 0

3 years ago

An electromagnetic wave of frequency 2.30 × 10^14 Hz propagates in carbon tetrachloride with a speed of 2.05 x 10^8 m/s. What is

KengaRu [80]

1. $8.91\cdot 10^{-7} m$

The wavelength of a wave is given by the formula

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

For the electromagnetic wave in this problem,

$f=2.30\cdot 10^{14}Hz$ is the frequency

$v=2.05\cdot 10^8 m/s$ is the speed of the wave

Substituting into the equation, we find

$\lambda=\frac{2.05\cdot 10^8 m/s}{2.30\cdot 10^{14}Hz}=8.91\cdot 10^{-7} m$

2. $22.1^{\circ}$

The angle of refraction can be found by using Snell's law:

$n_i sin \theta_i = n_r sin \theta_r$

where

$n_i = 1.00293$ is the refractive index of the first medium (air)

$n_r = 1.333$ is the refractive index of the second medium (water)

$\theta_i = 30.0^{\circ}$ is the angle of incidence in air

Solving the equation for $\theta_r$ , we find the angle of refraction of the light ray in water:

$\theta_r = sin^{-1} (\frac{n_i sin \theta_i}{n_r})=sin^{-1} (\frac{(1.00293)(sin 30^{\circ})}{1.333})=22.1^{\circ}$

5 0

3 years ago

A light ray is incident on a reflecting surface. If the light ray makes a "35°" angle with respect to the normal to the surface,

Tomtit [17]

the reflected light ray is 35° to the normal.

Explanation:

the light ray hitting the reflected surface is the incident ray, and forms an angle of incidence with the normal (the line perpendicular to the reflecting surface). The ray reflected fron the surface is called the reflected ray, and forms an angle of reflection.

since

the angle of incidence=angle of reflection

35° = 35°

6 0

3 years ago