The crew of an enemy spacecraft attempts to escape from your spacecraft by moving away from you at 0.259 of the speed of light.
1 answer:
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Answer:
v_average = 15 m / s
Explanation:
The average speed can be found in two ways,
* taking the distance traveled and divide it by the time spent
* taking the velocities in each time interval and then finding the weighted average by the time fraction
v_average = 1 / t_total ∑ vi ti
Let's apply this last equation
Total time is
t = t₁ + t₂
t = 10 + 10 = 20 min
v_average = 10/20 10 + 10/20 20
v_average = 10/2 + 20/2
v_average = 15 m / s
Answer:
Both of you did the same work but you expended more power.
Explanation:
<em>Work done</em> by an object is calculated by force applied multiplied by the distance.
W=F*d
From the figure given below let us calculate force applied bith you and yopur friend.
Let us take the stairs in positive x direction,
Work done by you W₁ ,
The force applied Fₓ = F - mgsinθ =maₓ
here aₓ = 0, because both of you move with constant speed
F - mgsinθ = 0
F= mgsinθ
The work done by you on the suitcase is
W = F L cos0° , where L is he length of the staircase.
W = FL = mgsinθL , by substituting value of F
Work done by you is W₁ = mgLsinθ
Similarly work done by your friend is W₂ = mgLsinθ.
Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .
Therefore <em>work done by both of you is same</em> . Both of you did equal work.
The power , is defined as amount of energy converted or transfered per second or rate at which work is done .
P = =
Power spend by you P₁ = mgLsinθ/t
P₁ = 15*9.8*Lsinθ/30
P₁ = 4.9L sinθ eqn 1
Power spend by your friend is P₂ = mgLsinθ/t
P₂ =15*9.8*Lsinθ/60
P₂ = 2.45Lsinθ eqn 2
Dividing eqn 1 and eqn 2
P₁ = 2P₂
You have spend more power than your friend .
Hence Both of you did equal work but you spend more power.
