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2 years ago

What is the medium the energy in the slinky wave travels through?

Physics

2 answers:

vlabodo [156]2 years ago

6 0

Answer:

The metal coil of the slinky

Explanation:

The wave travels on the actual slinky which we see deforming as the wave goes through.

Sergio [31]2 years ago

4 0

Answer:

A.) The metal coil of the slinky

Explanation:

Hello! This is the correct answer! Have a blessed day! :)

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2 years ago

Question 6 of 10

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1 year ago

If a certain mass of mercury has a volume of 0.002 m^3 at a temperature of 20°c, what will be the volume at 50°c

V125BC [204]

Change in volume = mass x coefficient of linear expansion x change in temperature
.002 x .0001802 x 30 = .000010812

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2 years ago

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WHAT HAPPENS TO THE VELOCITY AND ACCELERATION OF A BODY FALLING FREELY IN A VACUUM?

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both remains constant

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3 years ago

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

2 years ago