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3 years ago

How much to the left of the lens is the object?

Physics

1 answer:

stich3 [128]3 years ago

4 0

An object is 16.0 cm to the left of a thins lens. The lens forms an image 36.0 cm to the right of the lens.

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A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

elixir [45]

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

8 0

3 years ago

At which point is cool air circulating beneath warm air?

stepan [7]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters

Sphinxa [80]

Answer:

Long question good luck:)

Explanation:

3 0

3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

3 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 $\frac{\lambda}{2}$ .