Updated spacing between @ latex expressions

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правка #10274 предыдущая #9656 GitHub c702ebe ← раньше позже →
@@ -6,14 +6,14 @@
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<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
− <meta name="description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
+ <meta name="description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha . Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<meta name="author" content="Aliaksandr Melnichenka">
<meta name="date" content="2023-10" scheme="YYYY-MM">
− <meta property="og:title" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
+ <meta property="og:title" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha . Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
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− <meta property="og:description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
+ <meta property="og:description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha . Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
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− <title>Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.</title>
+ <title>Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha . Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.</title>
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