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Monica [59]
3 years ago
15

Kindly Only Answer if you know how to do this problem:

Mathematics
1 answer:
steposvetlana [31]3 years ago
7 0
Its representedf by 2 polygons(triangles)
first one :
24-20=4
11-6=5
second:
24-14=10
11-7=4
B. frind area of 2 small trianges
a=(base x height )/2
first=4x5/2=10ft   square
second=10x 4/2=20ft square
add them=10+20=30
rectangle24x 11=264
264-30=234
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A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739
OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

  1. For a constant acceleration: v_{f}=v_{0}+at, where  v_{f} is the final velocity in a direction after the acceleration is applied, v_{0} is the initial velocity in that direction  before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
  2. <em>Then for the x direction</em> it is known that the initial velocity is v_{0x} = 5320 m/s, the acceleration (the applied by the engine) in x direction is a_{x} 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the  is 739 s. Then: v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}
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3 years ago
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