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sp2606 [1]
3 years ago
10

A spaceship whose rest length is 350m has a speed of .82c

Physics
1 answer:
igomit [66]3 years ago
4 0

Answer:

t'=1.1897*10^{-6} s

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}

where:

v is the velocity of spaceship relative to certain frame of reference =  -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }

u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

t'=\frac{length}{Relatie seed (u')}

t'=\frac{350}{0.9806c}     (c = 3*10^8 m/s)

t'=\frac{350}{0.9806* 3.0*10^{8} }

t'=1.1897*10^{-6} s

t'=1.1897 μs

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A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will i
mafiozo [28]

Answer:

   t = 0.24 s

Explanation:

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation:  ΣF = ma

Rotation:      ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

                    ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation:     Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

         (1)                 mgSin27° - Ffr = ma  

For rotation:         (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

        (2)                Ffr = 2/3 ma

We substitute equation (2) in equation (1):

                            mgSin27° - 2/3 ma = ma

                            mgSin27° = ma + 2/3 ma

The mass gets cancelled:

                            gSin27° = 5/3 a

                            a = (3/5)(gSin27°)

                            a = (3/5)(9.8 m/s²(Sin27°))

                            a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

                           V = √2ad; where  d = 0.327m

                           V = √2(2.67 m/s²)(0.327m)

                            V = 1.32 m/s

Because V = d/t

                             t = d/V

                             t = 0.327m/1.32 m/s

                             t = 0.24 s

7 0
3 years ago
Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?
Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

x_{a}(t) = at+bt^2

x_{b}(t) = ct^2-dt^3

Constants are:

a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both x_a(t) and x_b(t) depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

at+bt^2 = ct^2-dt^3

2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0

t_1 = 4 - \sqrt{3} = 2.27 s, t_1 = 4 + \sqrt{3} = 5.73 s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0

t_1 = 1 s, t_2 = 4.33 s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67 s

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