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jeka94
3 years ago
9

A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is

a _______________.
a. Cassegrain focus
b. Newtonian focus
c. Schmidt-Cassegrain focus
d. Schmidt focus
Physics
1 answer:
erica [24]3 years ago
4 0
The correct answer for the question that is being presented above is this one: "Schmidt-Cassegrain focus." A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is a Schmidt-Cassegrain focus.
Here are the following choices: 
a. Cassegrain focus
b. Newtonian focus
c. Schmidt-Cassegrain focus
<span>d. Schmidt focus</span>
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A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x, t) = (5
barxatty [35]

Answer:

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?

4 0
3 years ago
Hi please may someone help me especially on the sketch part.
vaieri [72.5K]

Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.

We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s

we can use the formulae for acceleration to calculate the time taken/

(final - initial velocity)/timetaken=10

(30-0)/timetaken=10

timetaken =30/10=3 seconds

7 0
3 years ago
If the dartboard below is used to model an atom, which subatomic particles would be located at Z? *
Anna11 [10]

I know that protons and neutrons are located at the center of an atom, so the correct answer is D

7 0
3 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
Which describes refraction? *
sergey [27]

Answer:

Refracted rays travel through a boundary into a new medium.

Explanation:

Refracted rays travel through a boundary into a new medium. is only true for refraction.

The angle of incidence is the same for angle of refraction, is not true for refraction. Refraction follows Snell's law, states that ratio of the sine of the angle of refraction and the sine of the angle of incidence is always constant and equivalent to the ratio of phase velocities of the two mediums it is passing through.

Refracted rays change direction and go back to the original medium is false for refraction however, it is true for reflection.

5 0
3 years ago
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