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garri49 [273]
3 years ago
10

Three horses are side-by-side on a merry-go-round: one at the edge, one near the axis, and one in between. Each horse has the sa

me angular speed. Which horse has the greatest linear speed?
Physics
1 answer:
sineoko [7]3 years ago
4 0

Answer:

one at the edge

Explanation:

The relation between the linear velocity and the angular velocity is given by

v = r x ω

Where, v be the linear velocity, ω be the angular velocity and r be the radius of the circular path.

As the angular velocity is constant, thus, the linear velocity depends on the radius of circular path.

So, the horse which is near to the edge has maximum radius of circular path in which it is rotating. So, the horse which is at the edge of the merry go round has maximum linear speed.

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Answer:

B a measure of the amount of matter in an object

8 0
2 years ago
Can someone please help me ASAP!
Artist 52 [7]
I’m not 100% sure but i think it’s A because if you divide the speed by the time you get 2 and also all the other answer choices don’t make any sense!
3 0
3 years ago
Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri
Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part kx \pm \omega t, the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time \delta t (\delta t >0), the displacement <em>y</em> of the wave will be determined by the kx\pm\omega (t+\delta t) term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply \pm \omega (t+\delta t).

To know which side, right or left of the origin, would go through the origin after a time \delta t (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation \delta x (we would be looking where in space is what we would have in the future in time). The term would be then k(x+\delta x)\pm\omega t, which at the origin is k \delta x \pm \omega t. This would mean that, when the original equation has kx+\omega t, we must have that \delta x>0 for k\delta x+\omega t to be equal to kx+\omega\delta t, and when the original equation has kx-\omega t, we must have that \delta x for k\delta x-\omega t to be equal to kx-\omega \delta t

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when kx+\omega t, the part of the wave on the positive side (\delta x>0) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0
3 years ago
Which part of an atom has most of its mass? A. electrons B. neutrons C. nucleus D. protons
vredina [299]

<em>The </em><em>nucleus</em><em> has most of the atomic mass in an atom. The </em><em>nucleus</em><em> is made up of protons and neutrons.</em>

<em />

6 0
3 years ago
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Tju [1.3M]

Answer:

d. decreases

Explanation:

The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.

7 0
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