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Naya [18.7K]
3 years ago
10

Two male moose charge at each other with the same speed and meet on a icy patch of tundra. as they collide, their antlers lock t

ogether and they slide together with one-third of their original speed. what is the ratio of their inertias?
Physics
1 answer:
yanalaym [24]3 years ago
7 0
Their linear inertia is equivalent to their masses. Let the inertia of the first moose be m₁ and the second be m₂.

m₁u + m₂u = (m₁ + m₂) x 1/3 u
3m₁ + 3m₂ = m₁ + m₂
3 m₁/m₂ + 3 = m₁/m₂ + 1
m₁/m₂ = 2

The ratio of their inertias is 2
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3 years ago
Write the equation that links current, potential difference, and resistance
mrs_skeptik [129]

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==> Current = (potential difference) / (resistance)

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4 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
For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the
galina1969 [7]

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x =  52 cm * 1 / 100m = 0.52m

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So the frequency in the speed velocity is

f = 1 / T

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f ≅ 662 Hz

7 0
3 years ago
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