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3 years ago

A ________ is a government-regulated maximum price for goods.

1 answer:

8 0

D. price ceiling
This is a government regulation that establishes a maximum price for a particular good.

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Light travels at a constant speed of about 300,000 km/s. This speed is referred to as the speed of light. A light year is the di

Rama09 [41]

Based on the information in the question, a light year is the distance traveled light in one year. Then if the light from the star Centauri takes 4.2 years to reach the earth, then its distance from earth is 4.2 light years.

7 0

3 years ago

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What is one way that continental rifts are similar to mid-ocean ridges and what is one that they are different?

adell [148]

Explanation:

Continental rifts and mid-ocean ridges are both features of a divergent plate margin.

In both cases plates are moving away from one another. Therefore they are creating new land masses.

A continental drift like the east African rift valley is where a continent begins to pull apart or diverges.
A mid-ocean ridge is divergent margin in the ocean.

They are different in that, continental rift occurs within the continental plate that are on land.

But:

Mid-ocean ridges are in the oceanic crust in the ocean . They form the largest physiographic structure on the earth surface called the mid-ocean ridge.

learn more:

Descending lithosphere brainly.com/question/9582362

#learnwithBrainly

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3 years ago

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A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

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3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

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1 year ago

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A measure of randomness or disorder

vodka [1.7K]

It is simply called Entropy.

3 0

3 years ago

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