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2 years ago

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A spaceship is headed toward Alpha Centauri at 0.999c. According to us, the distance to Alpha Centauri is about 4 light-years. H

ow far away is Alpha Centauri according to the travelers in the ship?
a) also about 4 light-years
b) very slightly more than 4 light-years
c) very slightly less than 4 light-years
d) quite a bit less than 4 light-years
e) quite a bit more than 4 light-years

1 answer:

aniked [119]2 years ago

4 0

Answer:

According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>

<em></em>

Explanation:

For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:

v = d / t

where

v is the speed of the spaceship
d is the distance
t is the time

Therefore,

d = v × t

d = (0.999 c)(4 light-years)

d = 3.996 light-years

This distance is<em> very slightly less than 4 light-years. </em>

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AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

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Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

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ii) for upward motion of m1

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1 year ago

A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu

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Answer:

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2 years ago

A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath

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Answer:

A

Explanation:

Solution:-

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- Ignoring Doppler Effect.

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1 year ago

1. California sea lions can swim as fast as 40.0 km/h. Suppose a sea lion begins to chase a fish at this speed when the fish is

Pie

#1

In order to chase the fish the distance traveled by sea lion in time t must be equal to the distance of sea lion from the fish and distance traveled by fish in the same time.

So here we can say let say sea lion chase the fish in time "t"

then here we have

$d_1 = d_2 + L$

here

d1 = distance covered by sea lion in time t

d2 = distance covered by fish in the same time t

L = distance between fish and sea lion initially = 60 m

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$\frac{60}{9}*t = 60$

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So it will take 9 s to chase the fish by sea lion

# 2

velocity of truck on road = 25 m/s along North

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$v_{dt} = 1.43 \hat j + 1 \hat i$

we can write the relative velocity as

$v_d - v_t = 1.43 \hat j + 1 \hat i$

$v_d = v_t + (1.43 \hat j + 1 \hat i)$

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$v_d = 25 \hat j + (1.43 \hat j + 1 \hat i)$

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$v_d = \sqrt{26.43^2 + 1^2} = 26.44 m/s$

direction will be given as

$\theta = tan^{-1}\frac{v_x}{v_y}$

$\theta = tan^{-1}\frac{1}{26.43} = 2.2 degree$

so with respect to ground dog velocity is 26.44 m/s towards 2.2 degree East of North

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2 years ago

Read 2 more answers