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Vikentia [17]
3 years ago
12

A sound wave travels through air. The wavelength of the sound wave is equal to the distance between _____. adjacent centers of a

ir compression adjacent crests of the wave the source and observer of the wave adjacent troughs of the wave
Physics
2 answers:
ki77a [65]3 years ago
8 0

Adjacent crests of the wave

ololo11 [35]3 years ago
4 0

the sound wave is equal to the distance between the speed of sound

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Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra
malfutka [58]

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the  period of rotation of the mass is mathematically represented as

       T = 2 \pi \sqrt{\frac{I}{k} }

Here I is the moment of inertia of the mass about the rotation axis and  k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

6 0
3 years ago
How are density and pressure related related to each other?
lawyer [7]

Answer:

The pressure is the measure of force acting on a unit area. Density is the measure of how closely any given entity is packed, or it is the ratio of the mass of the entity to its volume. The relation between pressure and density is direct. Change in pressure will be reflected in a change in density

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3 years ago
A basketball is shot at 14.0 m/s at a 65.0 degree angle. What is the magnitude only (no direction) of the velocity of the ball 2
Alex787 [66]

Answer:

10.4 m/s

Explanation:

I just know it's right!

8 0
3 years ago
What is the main fuel consumed in the core of a red giant?<br> a. H<br> b. C<br> c. Fe<br> d. He
vladimir1956 [14]
<span>What is the main fuel consumed in the core of a red giant?
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4 0
3 years ago
A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-
joja [24]

Answer:

5.25\cdot 10^{40} kg m^2/s

Explanation:

The angular momentum of the pulsar is given by:

L=m\omega r^2

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r = 10.0 km = 1\cdot 10^4 m is the radius

\omega is the angular speed

Given the period of the pulsar, T=33.5\cdot 10^{-3} s, the angular speed is given by

\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s

And so, the angular momentum is

L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s

8 0
3 years ago
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