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rodikova [14]
3 years ago
9

You are flying your personal rocketcraft at 0.9c from Star A toward Star B. The distance between the stars, in the stars' refere

nce frame, is 1.0 ly. Both stars happen to explode simultaneously in your reference frame at the instant you are exactly halfway between them.
The question asked of the aforementioned scenario is "Do you see the flashes simultaneously?" and what is the time difference between the two?
Physics
1 answer:
Alexxx [7]3 years ago
7 0

Answer:

See explanation.

Explanation:

If both stars explode in simultaneously in the <em>your </em>frame of reference then obviously you will see the two flashes simultaneously, and therefore, the time difference between the events would be zero.

If however, the stars exploded simultaneously in their frame of reference, then you would not observe the flashes simultaneously.  Then the time difference between the events will not be zero, rather, you will observe star B exploding first and star A after.

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Let T be a linear transformation from a vector space V with dimension 11 onto a vector space W with dimension 7. What is the dim
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Answer:

The dimension of the nullspace of T = 4

Explanation:

The rank/dimension theorem is explains that:

Suppose V and W are vector spaces over F, and T:V → W is linear. If V is finite dimensional, then

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rank(T) = dimension of T = dim(T) = dim(W) = 7

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8 0
3 years ago
An electron is accelerated from rest by a potential difference of 412 V. It then enters a uniform magnetic field of magnitude 18
Salsk061 [2.6K]

Explanation:

Given that,

Potential difference, V = 412 V

Magnitude of magnetic field, B = 188 mT

(a) The potential energy of electron is balanced by its kinetic energy as :

eV=\dfrac{1}{2}mv^2

v is speed of the electron

v=\sqrt{\dfrac{2eV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 412}{9.1\times 10^{-31}}} \\\\v=1.2\times 10^7\ m/s

(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :

r=\dfrac{mv}{eB}\\\\r=\dfrac{9.1\times 10^{-31}\times 1.2\times 10^7}{1.6\times 10^{-19}\times 188\times 10^{-3}}\\\\r=3.63\times 10^{-4}\ m

Hence, this is the required solution.                                

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Which type of energy is transferred from one object to another by simply machines
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Kinetic Energy

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