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Brums [2.3K]
3 years ago
14

A spacecraft is traveling in interplanetary space at a constant velocity. What is the estimated distance traveled by the spacecr

aft in 9.50 seconds?
A) 2.30×10^2
B) 2.42×10^2
C) 2.51×10^2
D) 3.02×10^2
Physics
2 answers:
nalin [4]3 years ago
7 0

Answer:

2.30 × 102 kilometers

Explanation:

PLATO

vekshin13 years ago
3 0

i would say that the correct answer would be cc


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A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running
Katena32 [7]

Answer:

1

 t_a  =  13.49 \  s

2

 The distance is   D = 55.2 \  m    

Explanation:

From the question we are told that

   The maximum speed of the cheetah is  v =  103 \  km/h =  28.61 \  m/s

    The maximum of  gazelle is  u =  76.5 \  km/h =  21.25 \  m/s

     The distance ahead is d =  99.3 \  m

Let  t_a denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

           v* t_a = d +  u* t_a

=>          28.61 t_a = 99.3 +  21.25t_a

=>          7.36 t_a = 99.3

=>         t_a  =  13.49 \  s

Now at t =  7.5 s  

             7.5 v = D+  7.5u

=>          28.61 * 7.5  = D +  21.25* 7.5

=>          7.36 *  7.5 =D

=>         D = 55.2 \  m        

Hence the for the gazelle to escape the cheetah it must be 55.2 m

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Please help with this
sertanlavr [38]

Answer:I have to say 56

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