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bezimeni [28]
1 year ago
11

A rocket train car that is 30 m long is traveling from Los Angeles to New York at v=0.5c when a light at the center of the car f

lashes. When the light reaches the front of the car, it immediately rings a bell. Light reaching the back of the car immediately sounds a siren. A bicyclist waiting to cross the tracks sees the light in front of her at the instant when it flashes. Choose frame S' so that its origin is located at the center of the train. Take the direction of motion of the train (eastward) to be the positive direction of x'. Let the instant when the light flashes be t′=0s.
Required:
What are (x′A,t′A) and (x′B,t′B), the spacetime coordinates of event A and B in S'?
Physics
1 answer:
Nataly [62]1 year ago
5 0

Answer: The reference frame of a passenger in a seat near the center of the train

Explanation:

the speed of light is the same for the passenger and the bicyclist

then the avents are simultaneous fo the passenger not for the bicyclist

the delay between the two events for the bicyclist is

Δt=Δd/vs

where

Δd= lenght of train

vs=speed of sound

the reference frame of a passenger in a seat near the center of the train

Solution:

The space and time transformations are:

x' = γ(x - vt)

t' = γ(t - vx/c²).

In the primed frame the two events are simultaneous, so that Δt' = 0. Also here Δx' = 30. In the unprimed frame Δx' = 30 = γ(Δx - v Δt).......(*)

We also have Δt' = 0 = γ(Δt - vΔx/c²)→Δx = c²Δt/v......(**)

Substituting (**) in (*): 30 = γ(c²Δt/v - vΔt))→Δt = 30/(c²/v - v) =

30/(2c - 0.5c) = 6.7 x 10^(-8)s

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docker41 [41]

Answer:

Magnetic field at the center of the loop B=5.89\times 10^{-5}\ T.

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.

We know , magnetic field at the center of loop is given by :

B=N\dfrac{\mu_o i}{2r}

Putting all values in above equation we get :

B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.

Hence , this is the required solution.

8 0
1 year ago
I take 1.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium is reached, I have 2.0 kg of ice at 0°C. The water w
VashaNatasha [74]

Answer:

.c. −160°C

Explanation:

In the whole process one kg of water at  0°C loses heat to form one kg of ice and heat lost by them is taken up by ice at −160°C . Now see whether heat lost is equal to heat gained or not.

heat lost by 1 kg of water at  0°C

= mass x latent heat

= 1 x 80000 cals

= 80000 cals

heat gained by ice at −160°C to form ice at  0°C

= mass x specific heat of ice x rise in temperature

= 1 x .5 x 1000 x 160

= 80000 cals

so , heat lost = heat gained.

5 0
1 year ago
A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of
olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through 90^{o} then position of two mirrors will be changed. The resultant path difference will be as follows.

      \frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}

Formula for change in fringe shift is as follows.

          n = \frac{2lv^{2}}{\lambda c^{2}}

       v^{2} = \frac{n \lambda c^{2}}{2l}

             v = \sqrt{\frac{n \lambda c^{2}}{2l}}

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

             l = 11 m

    \lambda = 5.9 \times 10^{-7} m

           c = 3.0 \times 10^{8} m/s

Hence, putting the given values into the above formula as follows.

            v = \sqrt{\frac{n \lambda c^{2}}{2l}}

               = \sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}

               = 2.41363 \times 10^{9} m/s

Thus, we can conclude that velocity deduced is 2.41363 \times 10^{9} m/s.

3 0
1 year ago
Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what meth
Svetradugi [14.3K]

Answer:

If they are metallic spheres  they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

8 0
2 years ago
Alan wrote the following examples of changes in substance.
Yuliya22 [10]
A campfire being lighted and plants converting carbon-dioxide and water into glucose and oxygen are both forms of chemical change.

Therefore, the answer is:

B. Both are examples of chemical change.
5 0
2 years ago
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