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Vera_Pavlovna [14]
3 years ago
11

A package falls out of an airplane that is flying in a straight line at a constant altitude and speed. If you ignore air resista

nce, what would be the path of the package as observed by the pilot? As observed by a person on the ground?
Physics
1 answer:
Anettt [7]3 years ago
3 0

Answer:

Explanation:

If you ignore air resistant, then nothing affects the package horizontal motion. In Newton's first law it would keep the package at a constant speed, the speed of the airplane.

So to the eyes of the pilot who is also moving at the same horizontal speed, the lateral position of the package does not change. He can only perceive that the package is getting further away from him as it's dropping vertically.

To a person on the ground then the package is travelling in a parabolic path, where its horizontal speed is constant but vertical speed is increasing toward the ground at the rate of g.

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Using the force table, components of a vector can be found experimentally by suspending masses from 2 orthogonal strings which o
SCORPION-xisa [38]
Answer: 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).

Explanation:

1) Since the mass of <span>155 g is suspended at 210 degrees, you need to find the components of its weight on the orthogonal coordinate system (0° and 90°).
</span>

<span>2) You do that using the trignometric ratios sine and cosine.
</span>

<span>Weight is mass × g.
</span>
<span>Weight of the object = 155g × g
</span>
<span>Angle, α = 210°
</span>

<span>Horizontal component (0°)
</span>
<span>cosα = horizontal / hypotenuse ⇒ horizontal = hypotenuse × cosα
</span>
⇒ horizontal = 155g × g × cos(210°) = - 134.23g  × g

Vertical component
sinα = vertical / hypotenuse ⇒ vertical = hypotenuse × sinα
⇒ vertical = 155g × g × sin(210°) = -77.5g × g

3) Conclusion:

Therefore, the masses that must be suspended to balance the forces of the 155g mass are 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).




8 0
3 years ago
Displacement is the direction and distance an object is in motion<br><br> True or False
stellarik [79]
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A car starts from rest and after 10 seconds is traveling at 20 m/s. Assuming that it continues to accelerate at the same rate it
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Assuming that it continues to accelerate at the same rate it will take another 10 seconds to reach 40 m/s.

Answer:

Explanation:

Since the first question states that there is a change in the velocity from rest to 20 m/s in 10 seconds time interval. So the acceleration experienced by the car during this 10 seconds should be determined first as follows:

Acceleration = (final velocity-initial velocity)/Time

Acceleration = (20-0)/10 = 2 m/s².

So this means the car is traveling with an acceleration of 2 m/s².

As it is stated that the car continues to move with same acceleration, then in the second case, the acceleration is fixed as 2 m/s², initial velocity as 20 m/s and final velocity as 40 m/s. So the time taken for the car to reach this velocity with the constant acceleration value will be as follows:

Time = Change in velocity/Acceleration

Time = (40-20)/2 = 20/2=10 s

So again in another 10 seconds by the car to reach 40 m/s from 20 m/s. Similarly the car will take a total of 20 seconds to reach from rest to 40 m/s value for velocity.

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