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Serggg [28]
3 years ago
15

People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T

he nearest star is 4.1×1016m away. Suppose you had a spacecraft that could accelerate at 1.3 g for half a year, then continue at a constant speed. (This is far beyond what can be achieved with any known technology.) Part A How long would it take you to reach the nearest star to earth?
Physics
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

The travel would take 6.7 years.

Explanation:

The equation for an object moving in a straight line with acceleration is:

x = x0 + v0 t + 1/2a*t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

In a movement with constant speed, a = 0 and the equation for the position will be:

x = x0 + v t

where v = velocity

Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74  m/s²:

Seconds in half a year:

1/2 year = 1.58 x 10⁷ s

x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)²  = 1.59 x 10¹⁵ m

Now let´s see how much time it takes the travel to the nearest star after this half year.

The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²

v = v0 + a t

Since the spacecraft starts from rest, v0 = 0

v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s

Using the equation for position:

x = x0 + v t

4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t

(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t

t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.

The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.

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Answer:

a) The resulting angular speed of platform is 1.38 rev/sec

b) The change in kinetic energy of the system is 53 J.

Explanation:

This question is incomplete. The complete question will be:

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to  7.0 k g .m 2  

a) What is the resulting angular speed of the platform? Answer in units of r e v / s .

b)What is the change in kinetic energy of the system? Answer in units of J.

<h3>ANSWER:</h3>

a)

we know that:

Angular Momentum = L = Iω

From conservation of momentum:

Lo = Lf

(Io) (ωo) = (If) (ωf)

ωf = (Io) (ωo)/(If)

ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)

<u>ωf = 1.38 rev/sec =</u>

b)

ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec

ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec

The kinetic energy for rotational motion is given as:

K.E = (1/2)Iω²

Thus, the change in kinetic energy will be:

ΔK.E = (K.E)f - (K.E)o

ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²

ΔK.E = (1/2)(Ifωf² - Ioωo²)

ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²

<u>ΔK.E = 53 J</u>

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