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saw5 [17]
3 years ago
14

Which describes what an outside observer at rest would observe about a spaceship that speeds up and approaches the speed of ligh

t?
Time moves faster on the spaceship, and the length of the spaceship decreases. Time moves faster on the spaceship, and the length of the spaceship increases. Time moves more slowly on the spaceship, and the length of the spaceship decreases.
Time moves more slowly on the spaceship, and the length of the spaceship increases.
Physics
2 answers:
Rzqust [24]3 years ago
7 0

Answer:

The correct answer choice is C.  

"Time moves more slowly on the spaceship, and the length of the spaceship decreases."

Explanation:

Answer on Q'let

Art [367]3 years ago
6 0
The outside observer, at rest relative to the spaceship, would see the spaceship
get shorter. and the clocks on the spaceship run slower than they should.

At the same time, the crew of the spaceship, looking back at the observer on
Earth, would see the observer on Earth get shorter, and the observer's clock
run slower than it should.

They would both be measuring what they see correctly.
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When a body falls to the earth, the earth also move up to meet it but the earth motion is not noticeable. why?
viva [34]
It is because earths mass is very large so its movement isnt noticibale
7 0
3 years ago
The way matter moves in a longitudinal wave
serg [7]

Answer:

A longitudinal wave is a type of mechanical wave, or wave that travels through matter, called the medium. In a longitudinal wave, particles of the medium vibrate in a direction that is parallel to the direction that the wave travels. Places where particles of the medium crowd closer together are called compression's.

Explanation:

6 0
4 years ago
A 2 kg ball is thrown down with 50J of energy from a height of 10m, what is its velocity before it strikes the ground (neglect a
pochemuha

Answer:

V=14

Explanation:

PE=KE

mgh=1/2mv^2

2(9.8)10=1/2(2)v^2

(radical) 196= (radical)v

V=14

7 0
3 years ago
A motorcycle is travelling at a constant velocity of 90 km/h (no acceleration). How many
Mazyrski [523]
2km*1000=2000m
S=ut+1/2at^2
a=0,thus s=ut
2000=90t
t=2000/90
=22.22s(to 4 s.f)
Please mark as brainliest
8 0
3 years ago
A model rocket blasts off and moves upward with an acceleration of 12 m/s2 until it reaches a height of 26 m. at that height, it
Diano4ka-milaya [45]

initial acceleration of rocket is given as

a = 12 m/s^2

h = 26 m

now we can use kinematics to find its speed

v_f^2 - v_i^2 = 2 a h

v_f^2 - 0 = 2 * 12*26

v_f = 24.98 m/s

now after this it will be under free fall

so now again using kinematics

v_f = 0

at maximum height

v_f^2 - v_i^2 = 2 a s

0 - 24.98^2 = 2 * (-9.8)* h

h = 31.8 m

total height from the ground = 31.8 + 26 = 57.8 m

Part b)

now after reaching highest height it will fall to ground

So in order to find the speed we can use kinematics again

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2*9.8*57.8

v_f = 33.67 m/s

Part c)

first rocket accelerate to reach height 26 meter and speed becomes 24.98 m/s

now we have

v_f - v_i = a t

24.98 - 0 = 12*t_1

t_1 = 2.1 s

after this it will reach to highest point and final speed becomes zero

v_f - v_i = at

0 - 24.98 = -9.8 * t

t_2 = 2.55 s

now from this it will fall back to ground and reach to final speed 33.67 m/s

now we have

v_f - v_i = at

33.67 - 0 = 9.8 * t

t_3 = 3.44 s

so total time is given as

<em>t = 3.44 + 2.55 + 2.1 = 8.1 s</em>

5 0
3 years ago
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