1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
What is the motion of the particles in this kind of wave?
Musya8 [376]
The answer to this question is B I think
7 0
3 years ago
Read 2 more answers
A baseball is launched horizontally from a height of 1.8 m. The baseball travels 0.5 m before hitting the ground. How fast is th
zysi [14]

Answer:

0.83 m/s

Explanation:

FIrst of all, we have to find the time of flight, i.e. the time the baseball needs to reach the ground. This can be done by using the equation for the vertical motion:

h=ut+\frac{1}{2}gt^2

where

h is the initial height

u = 0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Substituting h = 1.8 m and solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.8)}{9.8}}=0.61 s

We know that the horizontal distance travelled by the ball is

d = 0.5 m

Therefore, we can find the horizontal velocity (which is constant during the whole motion):

v= \frac{d}{t}=\frac{0.5}{0.60}=0.83 m/s

4 0
3 years ago
A car with two people traveling down the road has a mass of 100 kg and a velocity of 5 m/s. The car pulls over and picks up two
Andrej [43]

Answer:

New_momentum = 750 [kg.m/s]

Explanation:

Momentum is defined as

M  = mass* velocity

Original momentum

M_old = 100 Kg * 5 m/s = 500 [kg.m/s]

If the car pulls over and picks up two people. The mass of the car is now 150 kg, but the velocity stays the same.

New momentum

M_new = 150 Kg * 5 m/s = 750 [kg.m/s]

3 0
3 years ago
A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. Part A) At the instant
Gre4nikov [31]

A) 2.2\cdot 10^{-19} N

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I is the current in the wire

r is the distance from the wire

In this situation,

I = 5.20 A

r = 4.60 cm = 0.046 m is the distance of the electron from the wire

Therefore the magnetic field strength at the electron's location is

B=\frac{(4\pi \cdot 10^7)(5.20)}{2\pi (0.046}=2.26\cdot 10^{-5} T

The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by

F=qvB

where

q is the magnitude of the charge of the particle

v is its velocity

B is the magnetic flux density

For the electron of this problem,

q=1.6\cdot 10^{-19} C is the charge

v=6.10\cdot 10^4 m/s is the speed

B=2.26\cdot 10^{-5} T is the magnetic field

Substituting,

F=(1.6\cdot 10^{-19})(6.10\cdot 10^4)(2.26\cdot 10^{-5})=2.2\cdot 10^{-19} N

B) Same direction as the current in the wire

First of all we have to find the direction of the magnetic field lines, which are concentric around the wire. Assuming the wire carries a current pointing upward, then if we use the right-hand rule:

- The thumb gives the direction of the current -> upward

- The other fingers wrapped give the direction of the field lines -> anticlockwise around the wire (as seen from top)

Now the direction of the force can be found by using the right-hand rule. We have:

- direction of the index finger = direction of motion of the electron (toward the wire, let's assume from east to west)

- middle finger = direction of the magnetic field (to the north)

- Thumb = direction of the force --> downward

However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.

8 0
3 years ago
What is universal constant (G)
ahrayia [7]

☄ <u>Universal</u><u> </u><u>Gravitational</u><u> </u><u>Constant</u><u>(</u><u>G</u><u>)</u>

Gravitational constant is a constant of proportionality when F(Gravitational force) is proportional to product of masses and inversely proportional to the distance between them.

  • It can be defined as:- Universal gravitational constant is the magnitude of the force between a pair of 1 kg masses that are kept 1 metre apart.
  • The SI unit of G is N m²/kg² and the CGS unit of G is Dyne cm² / g².

<h3>☄ <u>Do</u><u> </u><u>you</u><u> </u><u>know</u><u>?</u></h3>

The value of G (universal gravitational constant) was found out by the scientist Henry Cavendish by using the sensitive torsion balance.

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
Read 2 more answers
Other questions:
  • An unlabeled hierarchical diagram of various astronomical bodies is shown below. The labels A, B, C, and D can be used to repres
    8·1 answer
  • Need help on question b
    7·1 answer
  • You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3
    7·1 answer
  • A planet orbits a star, in a year of length 2.35 x 107 s, in a nearly circular orbit of radius 3.49 x 1011 m. With respect to th
    12·1 answer
  • Know the parts of an animal cell and rheir function​
    9·1 answer
  • PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
    12·2 answers
  • A 3.8kw elective motor powers an inclined conveyer belt. It is designed to lift heavy boxes from the warehouse floor to loading
    6·1 answer
  • What's a beam of light?<br>​
    8·1 answer
  • Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in
    11·2 answers
  • What is unusual about the results of mass determinations of clusters of galaxies?
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!