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Oliga [24]
2 years ago
10

A simple harmonic oscillator has an amplitude of 3. 50 cm and a maximum speed of 26. 0 cm/s. What is its speed when the displace

ment is 1. 75 cm?.
Physics
1 answer:
Marianna [84]2 years ago
3 0

Answer:

1.) A simple harmonic oscillator has an amplitude of 3.50 cm and a maximum speed of 26.0 cm/s. What is its speed when the displacement is 1.75 cm? 2.) Both pendulum A and B are 3.0 m long. The period of A is T. Pendulum A is twice as heavy as pendulum B. What is the period of B? 3.) The time for one cycle of a periodic process is called the _ ? 4.) In simple harmonic motion, the acceleration is proportional to? 5.) The position of a mass that is oscillating on a spring is given by x= (18.3 cm) cos [(2.35 s-1)t]. What is the frequency of this motion?

Explanation:

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vladimir2022 [97]
It goes twice as fast as the first one. Twice
6 0
3 years ago
What is the equation for potential energy?
Mrrafil [7]
You can calculate potential energy by:
U = m.g.h

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Hope this helps!
7 0
3 years ago
Read 2 more answers
An x-ray beam of wavelength 1.4×10−10m makes an angle of 20° with a set of planes in a crystal(the Bragg angle)causing first ord
Gnom [1K]

Answer:

Second order line appears at 43.33° Bragg angle.

Explanation:

When there is a scattering of x- rays from the crystal lattice and interference occurs, this is known as Bragg's law.

The Bragg's diffraction equation is :

n\lambda=2d\sin\theta      .....(1)

Here n is order of constructive interference, λ is wavelength of x-ray beam, d is the inter spacing distance of lattice and θ is the Bragg's angle or scattering angle.

Given :

Wavelength, λ = 1.4 x 10⁻¹⁰ m

Bragg's angle, θ = 20°

Order of constructive interference, n =1

Substitute these value in equation (1).

1\times1.4\times10^{-10} =2d\sin20

d = 2.04 x 10⁻¹⁰ m

For second order constructive interference, let the Bragg's angle be θ₁.

Substitute 2 for n, 2.04 x 10⁻¹⁰ m for d and 1.4 x 10⁻¹⁰ m for λ in equation (1).

2\times1.4\times10^{-10} =2\times2.04\times10^{-10} \sin\theta_{1}

\sin\theta_{1} =0.68

<em>θ₁ </em>= 43.33°

4 0
3 years ago
A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below t
bezimeni [28]

Answer:

the gauge pressure at the upper face of the block is 116 Pa

Explanation:

Given the data in the question;

A cubical block of wood, 10.0 cm on a side.

height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m

density ρ = 790 kg/m³

Using expression for the gauged pressure;

p-p₀ = ρgh

where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.

we know that, acceleration due to gravity g = 9.8 m/s²

so we substitute

p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m

= 116.13 ≈ 116 Pa

Therefore, the gauge pressure at the upper face of the block is 116 Pa

4 0
3 years ago
Mr. Galonski loves to use the Electromagnetic Spectrum. Create a scenario in which Mr. Galonski is using waves from the electrom
Anestetic [448]

Answer:

The Scenario:

On a normal Sunday afternoon Mr. Golanski is sitting in his living room reading his book. He decides after a while to turn on the Television to see what’s on the news, (Mr. Golanski is using Radio waves when he turns his television on by signaling the TV from his remote control). After a few hours Mr. Golanski decides it’s time to have dinner. He heats up a quick meal in his microwave because he doesn’t have the patience for cooking. (He is using microwave radiation to heat his food because water molecules in food absorb the radiation). He sits down for his meal, and halfway through he starts to choke! In a panicked frenzy he runs to his bathroom to try and dislodge the obstacle from his throat. By doing so he switched on the fluorescent lights in his bathroom exposing himself to small amounts of ultraviolet radiation. (Fluorescent lights absorb UV radiation and transmit visible light along with small amounts of UV light). Unable to dislodge the obstacle from his throat Mr. Golanski seeks help from his neighbor who drives him straight to the ER. To treat him properly the physicians opt for a fluoroscopy to examine Mr. Golanski’s esophageal tract. (Thus he is making use of X-ray imaging to obtain a visual of his internal esophageal structure to check for the obstruction). Once treated and discharged from the hospital Mr. Golanski returns home grateful to have survived this ordeal with minimum damage.

Explanation:

The Electromagnetic Spectrum is the range of frequencies and wavelengths for different light waves. They range with increasing frequency from Radio waves, to Microwaves, to Infrared waves, to Visible waves, to Ultraviolet waves, to Infrared waves, to X-rays, and to Gamma rays. Several of which we use in our daily lives such as Radio waves when operating our television or using our cellular phones. We also use microwaves to heat our food or for communication with satellites. We are also exposed to natural Ultraviolet radiation from the sun; however, we can also get exposed to other forms such as from certain types of light bulbs. We see visible light in the form of all the colors we can detect around us. We make use of x-rays for imaging techniques widely used in medicine for diagnostics, as well as Infrared waves in our home security systems. The electromagnetic spectrum is always used as a part of our everyday life.

7 0
3 years ago
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