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3 years ago

Can anyone please tell me where these labels would go?

Physics

2 answers:

Nadya [2.5K]3 years ago

6 0

The least potential energy would go at the very bottom of the track. the greatest kinetic energy would be on the upper half of the track and the least kinetic energy would be on the lower half of the track. please review this on google if you are not sure.

Nataly_w [17]3 years ago

6 0

W is right but x will equal greatest amount of KE

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At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.

tatuchka [14]

<span>At the periphery of a hurricane the air is sinking, and several kilometers above the surface, in the eye, the air is sinking. </span>

6 0

3 years ago

You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T

monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

d = 1/600

d = 1.67 10⁻³ mm

d = 1-67 10⁻³ m

Let's calculate the angle

sin θ = m λ / d

θ = sin⁻¹ (m λ / d)

First order

θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₁ = sin⁻¹ (3.93 10⁻¹)

θ₁ = 23.14 °

Second order

θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₂ = sin⁻¹ (0.786)

θ₂ = 51.81 °

3 0

3 years ago

On the kelvin scale what is the freezing point of water

Alla [95]

Freezing point of the water is known as 273 K

Hope this helps!

8 0

3 years ago

On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

$T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s$

Angular velocity is given by

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s$

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m$

From Hooke's law

$mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2$

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0

2 years ago

A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A

Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength $\frac{\lambda}{2}$ .