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bezimeni [28]
3 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]3 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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an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b
Kay [80]

Answer:

333.3 m

Explanation:

Given

m =100g\ =\  0.1kg\\v = 100 m/s\\g = 10 m/s ^2

Potential energy =\frac{2}{3}\  of\  Kinetic\  energy......Equation(1)

We know that

Potential energy=mgh

Kinetic energy =\frac{1}{2} mv^{2}

Now From the Equation(1)

mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\  gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\  m

3 0
3 years ago
A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is
AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to W=mg (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to B=m_w g (m_w=mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

W'=W-B

which can be rewritten as

m'g = mg-m_w g

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

m_w = m-m'=540 g-342 g=198 g

(b) 1.98\cdot 10^{-4} m^3

If the rock is completely submerged, the volume of the rock corresponds to the volume of water  displaced.

The volume of water displaced is given by

V_w = \frac{m_w}{\rho_w}

where

m_w = 198 g = 0.198 kg is the mass of the water displaced

\rho_w = 1000 kg/m^3 is the density of the water

Substituting,

V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3

And so this is also the volume of the rock.

(c) 2727 kg/m^3

The average density of the rock is given by

\rho = \frac{m}{V}

where

m = 540 g = 0.540 kg is the mass of the rock

V=1.98\cdot 10^{-4} m^3 is its volume

Substituting into the equation, we find

\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3

3 0
3 years ago
What's the difference between a direct relationship and a positive one?
borishaifa [10]
A positive or direct relationship is one in which the two variables (we will generally call them x and y) move together, that is, they either increase or decrease together. In a negative or indirect relationship, the two variables move in opposite directions, that is, as one increases, the other descremases
5 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
fgiga [73]

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

8 0
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What is the term that describes the amount of water in a volume of air at a specific temperature compared to the amount of water
Natasha_Volkova [10]
I believe it is relative humidity.
6 0
3 years ago
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