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Oksi-84 [34.3K]
3 years ago
8

Suppose you are in a spaceship traveling at 99% of the speed of light past a long, narrow space station. Your direction of trave

l is parallel to the length of the station. If you measure lengths of objects on the station and also how time is passing on the station, what results will you get?
A) Lengths will appear shorter and time will appear to pass slower.
B) Lengths will appear shorter and time will appear to pass faster.
C) Lengths will appear shorter and time will appear to pass faster.
Physics
2 answers:
satela [25.4K]3 years ago
5 0

Answer:

A. Lengths will appear shorter and time will appear to pass slower.  

Explanation:

From theory of relativity, we know that:

L₀ = length of object, measured in stationary frame of reference

L = length of object measured from a frame moving with respect to object, called ‘relativistic length’

v = relativistic speed between observer and the object

c = speed of light

then,

L = L₀ √(1-v²/c² )

Hence, the length of the object decreases with the increase in its relativistic speed "v"

t₀ = time measured by clock at rest with respect to event.

t = time measured by clock in motion relative to the event

v = relativistic speed between observer and the object

c = speed of light

then,

t =  t₀/√(1-v²/c² )

Hence, the time increases with the increase in in relativistic speed and as a result it appears to pass slower.

Hence, the correct option is:

<u>A. Lengths will appear shorter and time will appear to pass slower.</u>  

 

iVinArrow [24]3 years ago
3 0

Answer:

B) Lengths will appear shorter and time will appear to pass faster.

Explanation:

This is in line with the laws of relativity.

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andrew-mc [135]
I think the answer would be C
7 0
3 years ago
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A 0.15 kg baseball is hit by a baseball bat. Right before it is hit, the ball’s speed is 30 m/s, and right after it is hit, the
Roman55 [17]

Answer:

Impulse will be 12 kgm/sec

So option (b) will be correct option

Explanation:

We have given mass of the baseball m = 0.15 kg

Ball speed before hit v_1=30m/sec

Ball speed after hitting  v_2=-50m/sec ( negative direction due to opposite direction )

We have to find the impulse

We know that impulse is equal; to the change in momentum

So change in momentum = m(v_1-v_2)=0.15(30-(-50))=0.15\times 80=12kgm/sec

So option (b) will be correct option

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2 years ago
Suppose a vector V makes an angle ϕ with respect to the y axis. Part A What could be the x and y components of the vector V ? Ex
mina [271]
Well i dont know how to explain sorry

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3 years ago
According to the graph above, how large of a force is needed in order to stretch the string 1.00 meters?
gulaghasi [49]

My guess for this one would be; 400 N

My reasoning would be; it starts at 0 on both X and Y, if you need to get to 1.00 meters thats 4/4. 1/4 of 1.00 is .25, and on .25 its on 100 so multiply it by 4 to make 1.00 and you get 400 N

7 0
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In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin
GuDViN [60]

Answer:

The velocity is v_b = 20.17 \ m/s

Explanation:

From the question we are told that

   The mass of the ball is  m = 0.245 \ kg

   The radius is  r =  59.8 \  cm  =  0.598 \ m

   The force is  F =  30.9 \ N

   The speed of the ball is  v = 16.0 \ m/s.

Generally the kinetic energy at the top of the circle is mathematically represented as

    K_t  =  \frac{1}{2} *  m  *  v^2

=> K_t  =  \frac{1}{2} *  0.245   *  16.0 ^2  

=> K_t  =  31.36 \ J  

Generally the work done by the force applied on the ball from the top to the bottom  is mathematically represented as

       W =  F *  d

Here  d is the length of  a semi - circular arc which is mathematically represented as

       d =  \pi * r

So

      W =  30.9 *  0.598

      W = 18.48 \ J

Generally the kinetic energy at the bottom is mathematically represented as

      K_b  =  \frac{1}{2} *  m *  v_b^2

=>    K_b  =  \frac{1}{2} *  0.245  *  v_b^2

=>   K_b  =  0.1225  *  v_b^2

From the law of energy conservation

     K_t +  W  =K_b

=>    31.36+  18.48 = 0.1225  *  v_b^2

=>    v_b = 20.17 \ m/s

3 0
3 years ago
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