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Delvig [45]
2 years ago
11

Suppose our Sun is about to explode. In an effort to escape, we depart in a spacecraft at v=0.800 c and head toward the star Tau

Ceti, 12.0 ly away. When we reach the midpoint of our journey from the Earth, we see our Sun explode, and, unfortunately, at the same instant, we see Tau Ceti explode as well.(a) In the spacecraft's frame of reference, should we conclude that the two explosions occurred simultaneously? If not, which occurred first?
Physics
1 answer:
Mars2501 [29]2 years ago
8 0

Sun exploded 16 years after Tau Ceti.

Considering a scenario in which the Sun is going to explode.

In order to escape the explosion, we depart in a spacecraft with a speed of v = 0.8c.

The star Tau Ceti is 12 life years away.

At the midpoint of the journey, the Sun as well as Tau Cetik explode. at the same instant.

Now, the distance between the Sun and the spacecraft:

L = L(p) √( 1 - (v/c)²)

L = (6 ly) √[1 - (0.8)²] = 3.6 ly

Now, we observe Sun to be flying away from us at a speed of 0.8c whereas the light from the Sun approaches at a speed 1c.  Therefore, the gap between the Sun and the blast of the Sun wave is opened at a speed of 1.8c.

Therefore, the time passed ever since the Sun exploded is:

t = 3.60 ly / 1.80c = 2 years.

Now, we observe Tau Ceti moving toward us at a speed of 0.8c, whereas the light from the blast approaches at a speed of 1c which is only 0.2c faster. Then the gap between the Tau Ceti and its blast wave is 3.60 ly and growing at a speed of 0.2c.

Therefore, the time for which it has been opening for is:

t = 3.60 ly/ 0.20c

t = 18 years.

Therefore, Tau Ceti exploded 16 years before the Sun.

Learn more about Sun here:

brainly.com/question/15837114

#SPJ4

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The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = \sqrt{400/3.9} \approx 10.12739367, at = v, so 3.9 * 10.12739367 \approx 39.5  m/s
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a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if
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Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
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The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

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  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

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False. False.

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A solar eclipse occurs only when the moon is new.

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