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mezya [45]
3 years ago
12

You are packing for a trip to another star. During the journey, you will be traveling at 0.99c. You are trying to decide whether

you should buy smaller sizes of your clothing because you will be thinner on your trip due to length contraction. You also plan to save money by reserving a smaller cabin to sleep in because you will be shorter when you lie down. Should you (a) buy smaller sizes of clothing, (b) reserve a smaller cabin, (c) do neither of these things, or (d) do both of these things?
Physics
1 answer:
Elenna [48]3 years ago
7 0

Answer:

Do neither of these things ( c )

Explanation:

For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula

L = (l) \sqrt{1 -\frac{v^{2} }{c^{2} } }

where l = Measured distance from object at rest, L =  contracted measured in relation to the observer , v = speed of clock , c = speed of light

you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .

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For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km
muminat

Answer with Explanation:

We are given that

y=(18-2x^2) km

x=(4t-3)km

Differentiate x and y w.r.t t

\frac{dx}{dt}=4

\frac{dy}{dt}=-4x\frac{dx}{dt}

\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)

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v_x=4

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Magnitude of velocity=\mid v\mid=\sqrt{v^2_x+v^2_y}

\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s

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a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0

a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64

Substitute t=1

a_x=0,a_y=-64

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\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2

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2 years ago
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Explanation:

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Explanation:

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A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along
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Answer:

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Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

F(x)=(3\ N/m^2)x^2+(6\ N/m)x

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

W=\int\limits {F{\cdot} dx}

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So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

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3 years ago
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