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mezya [45]
3 years ago
10

Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be loca

ted exactly midway between A’ and C’. At t’=0 a light flash occurs at B’ and expands outward as spherical wave.
1. According to an observer in S’, do the wave fronts arrive at A’ and C’ simultaneously?
2. According to an observer in S, do the wavefronts arrive at A’ and C’ simultaneously?
3. If you answered no to either 1. or 2., what is the difference in their arrival times and at which point did the front arrive first?
Physics
1 answer:
Inga [223]3 years ago
8 0

Answer:

1) an observer in B 'sees the two simultaneous events

2)observer B sees that the events are not simultaneous

3)  Δt = Δt₀ /√ (1 + v²/c²)

Explanation:

This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems

1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events

Consequently an observer in B 'sees the two simultaneous events

2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.

Consequently observer B sees that the events are not simultaneous

3) let's calculate the times for each event

        Δt = Δt₀ /√ (1 + v²/c²)

where t₀ is the time in the system S' which is at rest for the events

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A bicycle has a momentum of 23.4
Snowcat [4.5K]
  • Momentum=23.4kgm/s
  • Velocity=2m/s

\\ \rm\longmapsto Momentum=Mass\times velocity

\\ \rm\longmapsto Mass=\dfrac{Momentum}{Velocity}

\\ \rm\longmapsto Mass=\dfrac{23.4}{2}

\\ \rm\longmapsto Mass=11.7kg

4 0
3 years ago
Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier di
olga nikolaevna [1]
<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

length of V-1 = 49/50mm

= 0.98mm

so,

minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

☆ Therefore,

The minor measurement of the vernier scale is 0.02mm.

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

_♡_<em>mashi</em>_♡_

6 0
2 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is  acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g] 
D = (2V²sinθ cosθ)/g 
 D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2  or  θ = π/4


Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
3 years ago
The potential energy of a watermelon is 15.0 J. The watermelon is 3.0 m high. What is the mass of the watermelon?
maks197457 [2]

Answer:

m = 0.51[kg]

Explanation:

Potential energy is defined as the product of mass by gravity by height.

E_{pot}=m*g*h

where:

Epot = potential energy = 15 [J]

m = mass [kg]

g = gravity acceleration = 9.8 [m/s²]

h = elevation = 3 [m]

Now replacing:

E_{pot}=m*g*h\\15=m*9.8*3\\m = 0.51[kg]

5 0
3 years ago
A contact force exists between two interacting objects that physically contact each other. a non-contact force exists when two i
Murrr4er [49]
True.

A contact force is a force between two objects that are physically in contact with each other: an example of a contact force is the normal reaction of a table supporting a book.

A non-contact force is a force between two objects that are not physically in contact with each other: an example of non-contact force is the gravitational attraction between the Earth and the Moon.
4 0
2 years ago
Read 2 more answers
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