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White raven [17]

3 years ago

10

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star syste

m 14.4 light-years away (one light-year is the distance lighttravels in one year). Relative to a frame of reference that is fixed with respect to Earth, how longdoes it take Rob to complete the trip?

1 answer:

Slav-nsk [51]3 years ago

5 0

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

$1Ly =9.4605284*10^{15}m \rightarrow 'Ly'$ means Light Year

Then

$14.4Ly = 1.36231609*10^{17} m$

If we have that

$v= \frac{x}{t} \rightarrow t = \frac{x}{t}$

Where,

v = Velocity

x = Displacement

t = Time

We have that

$t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c$ = Speed of light

$t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}$

$t= 454105363 s (\frac{1hour}{3600s})$

$t= 126140 hours(\frac{1day}{24hours})$

$t= 5255.85 days(\frac{1 year}{365days})$

$t = 14.399 years$

Therefore will take 14.399 years

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Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d

xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

$F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F = 0.0111 \ N$

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

6 0

3 years ago

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

6 0

2 years ago

Someone help me asap!

bogdanovich [222]

Answer:the answer would be B, heavy rainfall causes a landslide on the side of the mountain.

Explanation:

Hydrosphere is all water or liquid elements in the earths atmosphere and geosphere is the solid parts, such as rock.

4 0

3 years ago

Read 2 more answers

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

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3 0

6 months ago

A stone is dropped from a bridge and hits the pavement below in two seconds. What is the velocity of the stone when it hits the

Helen [10]

We have: a = v/t
Here, t = 2 s [ Given ]
a = 9.8 m/s² [constant value for earth system ]

Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s

In short, Your Answer would be Option B

Hope this helps!

4 0

2 years ago

Read 2 more answers