How do you determine the minimum horizontal velocity
1 answer:
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Answer:
Acceleration, a = -31.29 m/s²
Explanation:
It is given that, Initial speed of the aircraft, u = 140 mi/h =
62.58 m/s
Finally, it stops, v = 0
Time taken, t = 2 s
Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :
a = t
0-62.58 2 a =
a = -31.29 m/s²
So, the acceleration of the aircraft is 31.29 m/s²
and the car is decelerating. Hence, this is the required solution.
(i found this answer becuase i could do the math)
Acceleration, a = -31.29 m/s²
Explanation:
It is given that, Initial speed of the aircraft, u = 140 mi/h =
62.58 m/s
Finally, it stops, v = 0
Time taken, t = 2 s
Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :
a = t
0-62.58 2 a =
a = -31.29 m/s²
So, the acceleration of the aircraft is 31.29 m/s²
and the car is decelerating. Hence, this is the required solution.
(i found this answer becuase i could do the math)
The final speed of the astronaut at the space station is <u>0.36 m/s</u> and the direction will be <u>opposite</u> to the direction of the tool kit thrown.
The mass of the astronaut = 80 kg
The mass of the tool kit = 15 kg
The speed of the space station = 1.25 m/s
The speed of the tool kit threw = 6 m/s
The final speed of the astronaut can be found using the conservation of momentum formula,
where m₁ is the mass of the astronaut
m₂ is the mass of the tool kit
and
is the speed of the space station
is the speed of the astronaut after throwing the toolkit
Let us substitute the known values in the above equation, we get
(80 x ) + (15 x 6) = (80 x 1.25) + (15 x 1.25)
(80 x ) + 90 = 100 + 18.75
80 x = 118.75 - 90
80 x = 28.75
= 28.75 / 80
= 0.36 m/s
Therefore, the final speed of the astronaut is <u>0.36 m/s</u>
By newton's third law, for every action, there will be an equal and opposite reaction. Thus, the direction of the astronaut's direction will be against the direction of the tool kit thrown.
Learn more about the conservation of momentum in
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