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ololo11 [35]
3 years ago
8

Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8 m/s2, which gives the illusion of normal gravi

ty during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0 × 108 m/s? (b) How far will it travel in so doing?
Physics
1 answer:
vovangra [49]3 years ago
4 0
(a).
It starts from rest, and its speed increases by 9.8 m/s every second.
One tenth the speed of light is (1/10) (3 x 10⁸ m/s) = 3 x 10⁷ m/s .
To reach that speed takes (3 x 10⁷ m/s) / (9.8 m/s²) = <u>3,061,224 seconds</u> .
That's about 35 days and 10 hours.

(b).
Distance traveled = (average speed) x (time of travel)
Average speed = (1/2) of (1/10 the speed of light) = 1.5 x 10⁷ m/s .
Time of travel is the answer to part (a) above.
Distance traveled = (1.5 x 10⁷ m/s) x (3,061,224 sec) = <u>4.59 x 10¹³ meters</u>
That's 45.9 billion kilometers.
That's 28.5 billion miles.
That's about 6.2 times the farthest distance that Pluto ever gets from the Sun.
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A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted
Snezhnost [94]
We can solve the problem by using Snell's law, which states 
n_i \sin \theta_i = n_r \sin \theta_r
where
n_i is the refractive index of the first medium
\theta_i is the angle of incidence
n_r is the refractive index of the second medium
\theta_r is the angle of refraction

In our problem, n_i=1.00 (refractive index of air), \theta_i = 28.0^{\circ} and n_r=1.63 (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
\sin \theta_r =  \frac{n_i}{n_r}  \sin \theta_r =  \frac{1.00}{1.63}  \sin 28.0^{\circ} = 0.288
From which we find
\theta_r = \arcsin (0.288)=16.7^{\circ}
6 0
3 years ago
A student carries a backpack for one mile. Another student carries the same backpack for two miles
Svetllana [295]

Answer:

Compared to the first student, the second student did twice as much work as the first student.

Explanation:

The work done by the first student will be equal to the Force exerted by the backpack on the student carrying it multiplied by one mile (Distance).  The work done by the second student will be equal to the Force exerted by the backpack on the student carrying it multiplied by two miles (Distance).

3 0
3 years ago
A 3.00 m-long 6.00-kg ladder pivoted at the top hangs down from a platform at the circus. A 42.0-kg trapeze artist climbs to a p
statuscvo [17]

Answer:

The period of the system of ladder and woman, T = 2.5 seconds

Explanation:

Mass of the ladder, m_1 = 6 kg

Mass of the artiste, m_2 = 42.0 kg

Length of the ladder, L = 42.0 kg

The total moment of inertia can be calculated using the equation:

I = \frac{1}{3} M_1 L^2 + m_2 (\frac{L}{2} )^2\\I = \frac{1}{3} *6*3^2 + 42* (\frac{3}{2} )^2\\I = 18 + 94.5\\I = 112.5 kg m^2

D = L/2 = 3/2

D = 1.5 m

The frequency of the system of ladder and woman follows that of a physical pendulum which can be given by the equation:

f = \frac{1}{2\pi } \sqrt{\frac{mgD}{I} } \\f = \frac{1}{2\pi } \sqrt{\frac{48*9.8*1.5}{112.5} }\\f = 0.4

The period of the system of ladder and woman is given by:

T = 1/f

T = 1/0.4

T = 2.5 seconds

5 0
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igor_vitrenko [27]

Answer:

Buwaya

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The rainy season in Philippines starts in the month of June the same time  when Buwaya (local name given to the box in Pegasus constellation) starts being seen in the sky after a brief period of disappearance from the sky.

7 0
3 years ago
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Which occurrence demonstrates dispersion?
katrin2010 [14]
A rainbow. Dispersion is the splitting of radiation into it's different wavelengths.
4 0
3 years ago
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