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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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Answer:

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Explanation:

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1 year ago

. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass

alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

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KE = 5(.5)(4kg)

KE = 10J

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2 years ago

One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

$m = m_{1}=m_{2}$

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Along X- axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}$

$v_{1}=u_{1}+u_{2}$

Put the value into the formula

$v_{1}=2.2i-0.80i$

$v_{1}=1.4i\ m/s$

Along Y-axis

$0=m_{1}v_{1}+m_{2}v_{2}$

$m_{1}v_{1}=-m_{2}v_{2}$

$v_{1}=-v_{2}$

Put the value into the formula

$v_{1}=-1.36j\ m/s$

Then the final speed of the first ball

$v_{1}=\sqrt{(1.4)^2+(1.36)^2}$

$v_{1}=1.95\ m/s$

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

$\tan\theta=\dfrac{v_{2}}{v_{1}}$

$\tan\theta=\dfrac{-1.36}{1.4}$

$\theta=\tan^{-1}\dfrac{-1.36}{1.4}$

$\theta=-44.16^{\circ}$

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0

3 years ago

Give at least one fact about subatomic particles

victus00 [196]

Answer:

- Particles smaller than atoms are called subatomic particles .

- There are three famous subatomic particles, proton, neutron and electron .

- The study of sub atomic particles are called particle physics

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6 0

2 years ago

A reconnaissance plane flies 605 km away from

kolezko [41]

Answer:

$v_{avg} =$ 355 m/s

Explanation:

Distance = 605 km

Initial speed = $v_{i}$ = 284 m/s

Final velocity = $v_{f}$ = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

$2as = v_{f}^{2}+v_{i}^{2}$

Then using first equation of motion, we find time

$v_{f} = v_{i}+at$

Then using the formula of average velocity, we find average velocity

$v_{avg}=\frac{total-distance}{total-time}$

Second method is very simple

$v_{avg}=\frac{v_{f}+v_{i} }{2}$

$v_{avg}=\frac{426+284}{2}$

$v_{avg} =$ 355 m/s

8 0

3 years ago