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MA_775_DIABLO [31]
1 year ago
12

A rod with a rest length of 1m whizzes past an observer at a speed of 0.995c, where c is the speed of light. Is it possible that

the observer could also measure the length of the moving rod to be 1 m? A. yes B. no
Physics
2 answers:
Tresset [83]1 year ago
6 0

Answer:

B. no

Explanation:

  • When any body moves at a speed comparable to the speed of light (<em>i.e. relativistic speed</em>) then the observer sees a contraction in length of the body along the axis of motion.

Assuming that the motion of the body is along the axis of the rod, then the observer will measure its length to be lesser than its length at rest.

<u>Then according to Einstein's theory of relativity:</u>

L=\frac{L_0}{\gamma}

where:

L_0=  original length of the object (along the direction of motion)

L = observed length of the rod

\gamma = Lorentz factor

\gamma\equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

abruzzese [7]1 year ago
6 0

Answer:

B. no

Explanation:

If any body moves at a rate equal to the speed of light (i.e. relativistic velocity), the observer experiences a contraction along the direction of motion in the span of the body.

Lets us suppose that the motion of the body is along the axis of the rod, the observer will measure the length of the rod which will be lesser than its length in rest position.

And contracted length is given by

L=L_o\sqrt{1-\frac{v^2}{c^2} }

Lo= original length of the object along its axis.

L= length measured by the observer.

v= 0.995c

c= speed of light

So,observer will measure a length less than 1 m

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Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 36.5 cm while travelin
tia_tia [17]

Answer:

f=8.219*10^{8}Hz

Explanation:

We are going to use the formula  v=fλ

Where v= velocity of radio waves

f= frequency

λ= wavelength of wave

  • radio waves are electromagnetic waves and as such they have the speed of light which is 3*10^{8}m/s.
  • also when a wave travels from one medium to another, the wavelength changes while the frequency remains the same.
  • calculating for the frequency of the wave in air also gives us the frequency in the window glass.

f=\frac{v}{λ}

v=3*10^{8}m/s.

λ=36.5 cm = 36.5/100= 0.365m

f=\frac{3*10^{8}m/s.}{0.365m}

f=8.219*10^{8}Hz

7 0
2 years ago
To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-tr
REY [17]

Answer:

Explanation:

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6 0
2 years ago
A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.
Anna007 [38]

Answer:

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

N - mg = ma

so normal force on the block is given as

N = mg + ma

now the magnitude of kinetic friction on the block is given as

F_f = \mu N

F_f = \mu (mg + ma)

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

a = - \frac{F_f}{m}

a = -\mu_k(g + a)

now we know that bag hits the opposite wall at L distance away in time t

so we have

d = v t + \frac{1}{2}at^2

L = vt - \frac{1}{2}(\mu_k)(g + a) t^2

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

8 0
1 year ago
A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.
myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

5 0
1 year ago
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