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tankabanditka [31]
3 years ago
8

A spaceship starting from a resting position accelerates at a constant rate of 9.8 meters per second per second. How long and ho

w far will it take the spaceship to reach a speed of 1 percent the speed of light (300,000,000 m/s)
HOW LONG ANSWER CHOICES
A. 3.5 days
B. 35.4 days
C. 354 days
D. 3403 days
HOW FAR ANSWER CHOICES
A. 3.1 × 106 m
B. 2.94 × 108 m
C. 7.8 × 1010 m
D. 4.6 × 1013 m
30 POINTS
Physics
1 answer:
Verizon [17]3 years ago
4 0

Accelerating at 9.8 m/s² means that every second, the speed is 9.8 m/s faster than it was a second earlier.  It's not important to the problem, but this number (9.8) happens to be the acceleration of gravity on Earth.

1% of the speed of light = (300,000,000 m/s) / 100 = 3,000,000 m/s .

Starting from zero speed, moving (9.8 m/s) faster every second,
how long does it take to reach  3,000,000 m/s ?

           (3,000,000 m/s) / (9.8 m/s²)  =  306,122 seconds .
                                                   (That's  5,102 minutes.)
                                                        (That's  85 hours.)
                                                     (That's  3.54 days.)

Speed at the beginning . . . zero .
Speed at the end . . . 3,000,000 m/s
Average speed . . . . . 1,500,000 m/s

Distance = (average speed) x (time)

               = (1,500,000 m/s) x (306,122 sec) = 4.592 x 10¹¹ meters

                                                                     =  459 million kilometers

                         That's like from Earth
                                                  to       Sun
                                                             to    Earth
                                                                    to        Sun. 

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DaniilM [7]
The correct answer would be to express large and small numbers.
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A car travels along a clear 10.0 km section of motorway in 6.0 minutes. It then drives through 3.0 km of roadwork in 3.0 minutes
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  • 6min=1/10h=0.1h
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\\ \bull\tt\dashrightarrow Avg\:Speed=\dfrac{Total\:Displacement}{Total\:Time}

\\ \bull\tt\dashrightarrow Avg\:Speed=\dfrac{10+3}{0.1+0.05}

\\ \bull\tt\dashrightarrow Avg\:Speed=\dfrac{13}{0.15}

\\ \bull\tt\dashrightarrow Avg\:Speed=86.6km/h

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In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude
melamori03 [73]
Answers:
(a) B_o  = 0.3466μT
(b) \lambda = 0.4488μm
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Explanation:
Given electric field(in y direction) equation:
E_y = 104sin(1.40 * 10^7 x -\omega t)

(a) The amplitude of electric field is E_o = 104. Hence

The amplitude of magnetic field oscillations is B_o =  \frac{E_o}{c}
Where c = speed of light

Therefore,
B_o =  \frac{104}{3*10^8} = 0.3466μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
\frac{2 \pi }{\lambda} = 1.40 * 10^7
\lambda =  \frac{2 \pi}{1.40} * 10^{-7}
\lambda =  0.4488μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = 6.68 * 10^{14}Hz


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