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bezimeni [28]
2 years 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]2 years 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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You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm
maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0
2 years ago
A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotati
fgiga [73]

Answer:

5.5 rad/sec

Explanation:

8 0
2 years ago
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How lon
frosja888 [35]

Answer:

t=0.704s

Explanation:

A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?

from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

V^{2} =u^{2} +2as

Vy^2=0+2*9.8*1.73sin40.5

Vy^2=22.021

Vy=4.69m/s

Vx^2=u^2+2*9.81*cos40.5

Vy^2=25.81

Vy=5.08m/s

V=(Vy^2+Vx^2)^0.5

V=47.71^0.5

V=6.9m/s

from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.

f=force applied

v=velocity final

u=initial velocity

m=mass of the toy, 0.046

f=ma

f=m(v-u)/t

v=u+at

6.9=0+9.8t

t=6.9/9.81

t=0.704s

4 0
2 years ago
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H
kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

5 0
2 years ago
A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0
2 years ago
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